1. Find the number of ways to choose a pair {a, b} of distinct numbers from the set {1, 2, . . . , 50} such that
(i) |a − b| = 5 (ii) |a − b| ≤ 5
2. There are 12 students in a party. Five of them are girls. In how many ways can these 12 students be arranged in a row if
(i) there are no restrictions?
JS The number of ways to arrange the 12 students is 12! = P12 12.
(ii) the 5 girls must be together (forming a block)?
JS The number of ways to arrange five girls within a block is 5!, while the number of ways to arrange 8 objects — 7 boys and 1 block of girls — is 8!, thus by M.P., the number of ways that the students can be arranged is 5! · 8!.
(iii) no 2 girls are adjacent? JS Given that H12
5 =
12−5+1
5 is the number of ways that five places can be chosen amongst
twelve, so that no two of the places are consecutive, which will be the positions for the girls, 5!ways to arrange the five girls amongst those positions and 7! ways to arrange the seven boys amongst the remaining positions, by M.P., the number of ways that the students can be arranged is 5! · 7! · 85.
(iv) between two particular boys A and B, there are no boys but exactly 3 girls?
JS There are 2! ways to arrange A and B, 53 ways to choose the three girls amongst the five that would be between A and B, 3! to arrange these three chosen girls, and 8! ways to arrange the block of five students (A, B and the three girls between them) with the remaining seven students. By M.P., the number of ways that the students can be arranged is 2!· 53 ·3!·8!. 3. m boys and n girls are to be arranged in a row, where m, n ∈ N. Find the number of ways this can
be done in each of the following cases: (i) There are no restrictions;
JS The number of ways to arrange m + n people in a row is (m + n)! = Pm+nm+n.
(ii) No boys are adjacent (m ≤ n + 1); JS There are Hm+n
m =
m+n−m+1
m ways of choosing m places among m + n such that no two
of these places are consecutive, which will be the positions of the boys, m! ways to arrange the boys among the chosen positions and n! ways to arrange the girls among the remaining positions, by M.P., the number of ways to arrange the students is m! · n! · n+1m .
(iii) The n girls form a single block;
JS There are n! ways to arrange the girls within their block and (m + 1)! ways to arrange m + 1objects — the m boys and the block of girls — by M.P., the number of ways to arrange the students is n! · (m + 1)!.
(iv) A particular boy and a particular girl must be adjacent.
JS There are 2! ways to arrange the boy and the girl within their block, and (m + n − 1)! ways to arrange m + n − 2 + 1 objects — the boy-and-girl block and the m + n − 2 other students — by M.P., the number of ways to arrange the students is 2! · (m + n − 1)!.
4. How many 5-letter words can be formed using A, B, C, D, E, F, G, H, I, J, (i) if the letters in each word must be distinct?
JS The number of arrangements of five distinct letters from among the ten letters is P10 5 = 10!
5! = 10 · 9 · 8 · 7 · 6.
(ii) if, in addition, A, B, C, D, E, F can only occur as the first, third of fifth letters while the rest as the second or fourth letters?
JS For the odd-numbered letters, the choices for the letters are 6, 5 and 4, respectively; for the even-numbered letters, the choices for the letters are 4 and 3, repectively. By M.P., the number of five-letter words that can be constructed is 6 · 4 · 5 · 3 · 4.
5. Find the number of ways of arranging the 26 letters in the English alphabet in a row such that there are exactly 5 letters between x and y.
6. Find the number of odd integers between 3000 and 8000 in which no digit is repeated. JS There are two types of numbers that satisfy the conditions:
Type 1: Numbers with odd thousands digits: the thousands digit must be 3, 5 or 7. This will leave only 4 possible digits for the ones digit, and 8 · 7 choices for the middle two digits. There are 3 · 8 · 7 · 4 = 672such numbers.
Type 2: Numbers with even thousands digits: the thousands digit must be 4 or 6. This will leave 5 possible digits for the ones digit, and 8 · 7 choices for the middle two digits. There are 2 · 8 · 7 · 5 = 560such numbers.
Thus, there are 672 + 560 = 1232 such numbers satisfying the conditions. 7. Evaluate
1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! where n ∈ N.
MS Let k = 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n!. Note that 1+k = 1! + 1 · 1! | {z } 2·1!=2! +2·2!+3·3!+· · ·+n·n! = 2! + 2 · 2! | {z } 3·2!=3! +3·3!+· · ·+n·n! = · · · = (n+1)·n! = (n+1)!. Thus k = 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! = (n + 1)! − 1. JS 1 · 1! + 2 · 2! + 3 · 3! + · · · + n · n! + 1! + 2! + 3! + · · · + n! 2! + 3! + 4! + · · · + (n + 1)! − 1! − 2! − 3! − · · · − n! − 1! + (n + 1)! 8. Evaluate 1 (1 + 1)!+ 2 (2 + 1)!+ · · · + n (n + 1)! where n ∈ N. MS Let k = 1 (1 + 1)!+ 2 (2 + 1)!+ · · · + n (n + 1)!. Note that 1 (1 + 1)!+ 2 (2 + 1)!+ · · · + n (n + 1)! + 1 (1 + 1)!+ 1 (2 + 1)!+ · · · + 1 (n + 1)! 2 (1 + 1)!+ 3 (2 + 1)!+ · · · + n + 1 (n + 1)! = 1 1!+ 1 2!+ · · · + 1 n! Thus k + 1 2!+ 1 3!+ · · · + 1 (n + 1)! = 1 1!+ 1 2!+ · · · + 1 n! and k = 1 2!+ 2 3!+ · · · + n (n + 1)! = 1 − 1 (n + 1)!. JS Prove, by induction, 1 (1 + 1)!+ 2 (1 + 1)!+ · · · + k − 1 k! = 1 − 1 k!: for k = 2: 1 2! = 1 2 = 1 − 1 2!.
induction: Assume that the formula holds for k = n, then 1 (1 + 1)!+ 2 (1 + 1)!+ · · · + n − 1 n! | {z } + n (n + 1)! = 1 − 1 n!+ n (n + 1)! = (n + 1)! − (n + 1) + n (n + 1)! = (n + 1)! − 1 (n + 1)! = 1 − 1 (n + 1)!.
9. Prove that for each n ∈ N,
(n + 1)(n + 2) · · · (2n) is divisible by 2n.
10. Find the number of common positive divisors of 1040and 2030.
JS Since 1040 = 240· 540 and 2030 = 260· 530, their common positive divisors are of the form
2a· 5b, where 0 ≤ a ≤ 40 and 0 ≤ b ≤ 30. Thus there is a bijection between the common positive
divisors of the numbers and the coordinate pairs (a, b) satisfying the given conditions; both sets have 41 · 31 = 1271 elements.
11. In each of the following, find the number of positive divisors of n (inclusive of n) which are mul-tiples of 3:
(i) n = 210;
JS n = 3 · 7 · 5 · 2, thus the each positive divisor of n which is a multiple of 3 is the product of 3 and a subset of the set {2, 5, 7}; thus, there are 30 + 3
1 + 3 2 +
3
3 = 8 positive divisors of
210 that are multiples of 3. (ii) n = 630;
JS n = 3 · 7 · 5 · 3 · 2, thus the each positive divisor of n which is a multiple of 3 is the product of 3 and a subset of the set {2, 3, 5, 7}; thus, there are 40 + 4
1 + 4 2 + 4 3 + 4 4 = 16 positive
divisors of 630 that are multiples of 3. (iii) n = 1512000.
JS n = 3 · 7 · 52· 32· 25, thus the each positive divisor of n which is a multiple of 3 is the product
of 3 and a multi-subset of the multi-set {5 · 2, 2 · 3, 2 · 5, 7}; thus, there are 6 · 3 · 3 · 2 = 108 positive divisors of 1512000 that are multiples of 3.
12. Show that for any n ∈ N, the number of positive divisors of n2is always odd.
KM Let n ∈ N with prime factorization n = p1k1p2k2· · · prkr. Then, n2 = p12k1p22k2· · · pr2kr. The
number of positive divisors of n2isQr
i=1(2ki+1)which is odd since (2ki+1)is odd and the product
of odd numbers is odd.
13. Show that the number of positive divisors of “111 . . . 1 | {z }
1992
” is even. TT, JA Using some basic divisibility rules, we know that 111 . . . 1111
| {z }
1992
is divisible by 3 but not 9. Thus, we can express 111 . . . 1111
| {z } 1992 as 111 . . . 1111 | {z } 1992 = 3r.
Let the prime factorization of r be r = pa1 1 p a2 2 · · · p ak k where a1, a2, . . . , ak, k ∈ N,
p1, p2, . . . , pkare prime numbers
Since 3r is not divisible by 9, r will not be divisible by 3. Hence, none of p1, p2, . . . , pk will actually
be a 3.
The prime factorization of 111 . . . 1111
| {z } 1992 is simply given by 111 . . . 1111 | {z } 1992 = 31pa1 1 p a2 2 · · · p ak k .
And thus the number of positive divisors that it has will be
We need not concern ourselves with the actual prime factors and their corresponding exponents, because as seen in the expression above, the product is always even.
Hence, the number of positive divisors for 111 . . . 1111
| {z }
1992
is always even. 14. Let n, r ∈ N with r ≤ n. Prove each of the following identities:
(i) Pn r = nP n−1 r−1, JS Pnr = n! (n − r)! = n (n − 1)! [(n − 1) − (r − 1)]! = nP n−1 r−1. (ii) Pn r = (n − r + 1)Pnr−1, JS Pn r = n! (n − r)! = n − r + 1 n − r + 1 n! (n − r)! = (n − r + 1) n! [n − (r − 1)]! = (n − r + 1)P n r−1. (iii) Pn r = n n − rP n−1 r , where r < n, JS Pnr = n! (n − r)! = n n − r (n − 1)! [(n − 1) − r]! = n n − rP n−1 r . (iv) Pn+1 r = Pnr+ rPnr−1, JS Pnr+ rPnr−1 = n! (n − r)!+ r n! [n − (r − 1)]! = n! · (n − r + 1) + r · n! (n − r + 1)! = n! · n − n! · r + n! + r · n! (n − r + 1)! = n! · (n + 1) [(n + 1) − r]! = n! · (n + 1) [(n + 1) − r]! = (n + 1)! [(n + 1) − r]! = Pn+1r . JS Pn+1
r is the number of arrangements of r distinct objects taken from n + 1 objects. This
number of arrangements may also be counted by considering two cases:
Case 1: The (n + 1)th object is not among the r distinct objects arranged. Therefore, the r objects must be from the n other objects, which has Pn
r arrangements.
Case 2: The (n + 1)th object is among the r distinct objects arranged. The (n + 1)th object can be placed in any of the r positions in the arrangement, and the remaining r − 1 distinct objects are taken from the n remaining objects. There are rPn
r−1such arrangements.
By A.P., the sum Pn
r + rPnr−1 is also the number of arrangements of r distinct objects taken
from n + 1 objects, and therefore Pn+1
r = Pnr+ rPnr−1. (v) Pn+1 r = r! + r(Pnr−1+ P n−1 r−1+ · · · + P r r−1). JS Using (iv), Pn+1 r = Pnr + rPnr−1 =⇒ Pn+1r − Pnr = rPnr−1 Pn r = Pn−1r + rP n−1 r−1 =⇒ Pnr − Pn−1r = rP n−1 r−1 Pn−1 r = Pn−2r + rP n−2 r−1 =⇒ P n−1 r − Pn−2r = rP n−2 r−1 .. . ... Pr+1 r = Prr+ rPrr−1 =⇒ Pr+1r − Prr = rPrr−1 Pn+1r = r! + r(Pnr−1+ Pn−1r−1+ · · · + Prr−1) ⇐= Pn+1r − Pr r = rP n r−1+ rP n−1 r−1 + · · · + rP r r−1
15. In a group of 15 students, 5 of them are female. If exactly 3 female students are to be students are to be selected, in how many ways can 9 students be chosen from the group
(i) to form a committee?
JA, JS From the given, we have 5 female students, thus, we have 10 male students, i.e., 15 − 5 = 10. Now, we have the following restrictions: that are exactly 3 female students are to be selected. So,
5 3
Since we need only exactly 3 students from the group of five female students and knowing that a committee is composed of 9 students, we have to choose 6 male students from the remaining 10 male students. For that case we have,
10 6
. (15b)
Hence, by M.P., multiplying (15a) and (15b) above, there are 5 3 10 6 = 2, 100
ways of choosing 9 students from to group of 15 to form a committee. (ii) to take up 9 different posts in a committee?
JA, JS To answer the second question: There are 9! ways for each students to take up different posts in a committee since every students in the committee could possibly take up these 9 position. Thus, by M.P., we have
5 3 10 6 9!.
16. Ten chairs have been arranged in a row. Seven students are to be seated in seven of them so that no two students share a common chair. Find the number of ways this can be done if no two empty chairs are adjacent?
JS There are 7! ways that the seven students can be arranged in a selection of seven chairs. At most one empty chair can be placed between any two chairs that seat students, and there may be an empty chair at the ends of the row, so there are eight possible places to have the three empty chairs, thus 83 ways to select the positions of the empty chairs in relation to the chairs that seat students. By M.P., the number of ways that the students can be seated is 7! · 83.
17. Eight boxes are arranged in a row. In how many ways can five distinct balls be put into the boxes if each box can hold at most one ball and no two boxes without balls are adjacent?
JS There are H8 3 =
8−3+1
3 ways to select three empty boxes among the eight boxes so that no
two empty boxes are next to each other, and 5! ways to arrange the balls among the non-empty boxes. By M.P., the number of ways to place the balls in the boxes are 5! · 63.
18. A group of 20 students, including 3 particular girls and 4 particular boys, are to be lined up in two rows with 10 students each. In how many ways can this be done if the 3 particular girls must be in the front row while the 4 particular boys be in the back?
JS There are 10! ways to arrange the students in the front row, 10! ways to arrange the students in the back row, and 136 ways to choose from the 20 − 3 − 4 students (other than the particular boys and girls) six other students to accompany the four boys in the back row. By M.P., the number of ways to line up the students is 10! · 10! · 136.
19. In how many ways can 7 boys and 2 girls be lined up in a row such that the girls must be separated by exactly 3 boys?
JS There are 2! to arrange the two girls, 73 ways of choosing the three boys who will be between the girls from among the seven boys, 3! ways to arrange the chosen boys between the girls, and (4 + 1)!ways to arrange the four remaining boys and the block of five people that have the girls at the ends. By M.P., the number of ways to line up the boys and girls is 2! · 73 · 3! · 5!.
20. In a group of 15 students, 3 of them are female. If at least one female student is to be selected, in how many ways can 7 students be chosen from the group
(i) to form a committee? LL, FO There are 31 12
5 ways to form the committee with exactly 1 female; there are 3 2
12 5
ways to form the committee with only two females; and there are 33 12
4 ways to form the
least one female is 31 12 6 + 3 2 12 5 + 3 3 12 4 = 5, 643.
LL, FO There are 157 ways to form the committee of 7 students with no restrictions and 12 7
ways to form the committee with no females. Thus the number of ways to form the committee with at least 1 female is 157 − 127.
(ii) to take up 7 different posts in a committee?
LL, FO Since there are 7! ways for each student to take up different posts in a committee and every student in the committee could possibly take the 7 positions. Thus, by MP, we have 7! 157 − 127 = (5040)(5643) = 28, 440, 720.
21. Find the number of (m + n)-digit binary sequences with m 0s and n 1s such that no two 1s are adjacent, where n ≤ m + 1.
22. Two sets of parallel lines with p and q lines each are shown in the following diagram:
p | {z } q Find the number of parallelograms formed by the lines.
MS Since a parallelogram is formed by two intersecting pairs of parallel lines, choosing two lines amongst the p horizontal lines and two lines amongst the q diagonal lines creates a parallelogram, and each parallelogram in the figure is created by a pair amongst the p horizontal lines and a pair amongst the q diagonal lines, the number of parallelograms in the diagram is equal to the number of ways that a pair of horizontal lines and a pair of diagonal lines can be selected, which is p2 q
2.
23. There are 10 girls and 15 boys in a junior class, and 4 girls and 10 boys in a senior class. A commit-tee of 7 members is to be formed from these 2 classes. Find the number of ways this can be done if the committee must have exactly 4 senior students and exactly 5 boys.
FO Restriction:
1) 7 members of the committee form 2 classes, 2) exactly 4 senior students,
3) exactly 5 boys. Consider the ff. cases:
Case 1: Since we only need 7 members of the committee from 2 classes having exactly 4 senior students and exactly 5 boys, we choose 4 boys from the senior class thus we write 104. Because the committee must have exactly 5 boys, we have to pick 7 boys from the junior class. We cannot pick another boy from the senior class since we are restricted that exactly 4 senior students must be in the committee and it must have exactly 5 boys. Thus we choose 1 boy from the junior class having 15 boys and write 151. Now, we need to choose 2 female students from the junior class to complete the committee, we have 102.
By (MP), 104 15 1
10
2 = 141, 750 ways of choosing 4 senior boys, 1 boy from the junior class
and 2 girls from the junior class. Case 2: Choosing
3 boys from the senior class 1 girl from the senior class 2 boys from the junior class 1 girl from the junior class By (MP), 103 4 1 15 2 10 1 = 504, 000 ways.
Case 3: Choosing
2 boys from the senior class 2 girl from the senior class 3 boys from the junior class By (MP), 102 4
2
15
3 = 122, 850 ways.
Therefore, by (AP), there are 141, 750 + 504, 000 + 122, 850 = 768, 600 ways a committee of 7 members formed from 2 classes having exactly 4 senior students and exactly 5 boys.
24. A box contains 7 identical white balls and 5 identical black balls. They are to be drawn randomly, one at a time without replacement, until the box is empty. Find the probability that the 6th ball drawn is white, while before that exactly 3 black balls are drawn.
PB 5 2 6 2 12 5 = 25 132.
Note that the first factor in the numerator is the number of ways of arranging the first five balls, three of which are black and two are white. The second factor in the numerator is the number of ways of arranging the last six balls four of which are black and two are white. Also, the sixth ball is fixed to be a white ball. The denominator is just the total number of arranging twelve balls, seven of which are black and five are white.
25. In each of the following cases, find the number of shortest routes from O to P in the street network shown below: r O rP r A rB rC
(i) The routes must pass through the junction A;
PB We first count the number of ways to go from point O to point A, then from point A to point P . This is given by:5
2 8
3
= 560. (ii) The routes must pass through the street AB;
PB We first count the number of ways to go from point O to point A, then from point B to point P . This is given by:5
2 7
3
= 350. (iii) The routes must pass through junctions A and C;
PB We first count the number of ways to go from point O to point A, then from point A to point C, then finally from point C to point P . This is given by: 5
2 4 1 4 2 = 240. (iv) The street AB is closed.
PB We first count the number of ways to go from point O to point P , then we subtract the number of ways we pass through AB. This is given by:13
5 −5 2 7 3 = 937.
26. Find the number of ways of forming a group of 2k people from n couples, where k, n ∈ N with 2k ≤ n, in each of the following cases:
(i) There are k couples in such a group; (ii) No couples are included in such a group; (iii) At least one couple is included in such a group; (iv) Exactly two couples are included in such a group.
27. Let S = {1, 2, . . . , n + 1} where n ≥ 2, and let
T = {(x, y, z) ∈ S3| x < z and y < z}. Show by counting |T | in two two different ways that
n X k=1 k2= |T | =n + 1 2 + 2n + 1 3 . KM
Method 1: If z ≥ 2, then there are z − 1 choices for both x and y. Thus, |T | =Pn+1
i=2(i − 1) 2=Pn
k=1k 2.
Method 2: Consider 2 disjoint cases.
Case 1: x = y. Then, there are n+12 such triples. Choose 2 numbers from S, the larger one is z, and the smaller one is x and y.
Case 2: x 6= y. Then, there are 2 n+13 such triples. First, choose 3 numbers from S, the largest one is z. We multiply by 2 because there are two possible positions for the remaining two numbers from S (after having set z).
Thus, |T | = n+12 + 2 n+1 3 .
28. Consider the following set of points on the xy-plane:
A = {(a, b) | a, b ∈ Z, 0 ≤ a ≤ 9 and 0 ≤ b ≤ 5}. Find
PB Note that we have 10 a coordinates, and 6 b coordinates. (i) the number of rectangles whose vertices are points in A;
PB To form a rectangle, we simply select two points from the 10 a coordinates, and two from the 6 b coordinates. This is given by:10
2 6
2
= 675. (ii) the number of squares whose vertices are points in A.
PB We count the number of squares using cases:
1 × 1square: 5 · 9 such squares 2 × 2square: 4 · 8 such squares 3 × 3square: 3 · 7 such squares 4 × 4square: 2 · 6 such squares 5 × 5square: 1 · 5 such squares Hence, there are 115 squares.
29. Fifteen points P1, P2, . . . , P15are drawn in the plane in such a way that besides P1, P2, P3, P4, P5
which are drawn collinear, no other 3 points are collinear. Find
(i) the number of straight lines which pass through at least 2 of the fifteen points;
PB We basically count the number of ways of connecting any two points. Now, since five points are collinear, then we subtract the number of ways of connecting any two of these five points, and add 1 for the line formed by these five collinear points. This is given by:
15 2 −5 2 + 1 = 96. (ii) the number of triangles whose vertices are 3 of the 15 points.
PB We basically count the number of ways of connecting any three points to form a triangle. Now, since five points are collinear, then we subtract the number of ways of connecting any three of these five points since those three will not form a triangle. This is given by:
15 3 −5 3 = 445.
30. In each of the following 6-digit natural numbers:
333333, 225522, 118818, 707099,
every digit in the number appears at least twice. Find the number of such 6-digit natural numbers. 31. In each of the following 7-digit natural numbers:
1001011, 5550000, 3838383, 7777777,
every digit in the number appears at least 3 times. Find the number of such 7-digit natural num-bers.
MS A 7-digit must begin with a nonzero digit, giving nine possible options. In relation to the first digit, there are three cases to consider:
Case 1: The first digit is repeated twice in the next six digits. The four remaining digits will be the same, so the possibilities of this case are determined by the location of the two repetitions of the first digit, and the choice of the remaining digit. Hence, this case has 9 · 62 · 9 elements. Case 2: The first digit is repeated thrice in the next six digits. The three remaining digits will be the
same, so the possibilities of this case are determined by the location of the three repetitions of the first digit, and the choice of the remaining digit. Hence, this case has 9 · 63 · 9 elements. Case 3: The first digit is repeated in the next six digits. This case has 9 elements.
So there are a total of 9 · 62 · 9 + 9 · 6
3 · 9 + 9 = 2844 such 7-digit natural numbers.
32. Let X = {1, 2, 3, . . . , 1000}. Find the number of 2-element subsets {a, b} of X such that the product a · bis divisible by 5.
PB Any product of two numbers is divisible by 5 if at least one of the two numbers is a multiple of 5. Thus the product of (5k)(a) is divisible by 5 for any natural numbers a and k. Let us now answer the problem. The 2-element subsets we are looking for is in the form {5k, a} (order does not matter). Thus, the smallest value of k is 1 and the largest is k = 200. Thus we consider 200 cases.
Case 1. Let k = 1, thus {5k, a} = {5, a}. There are 999 subsets of this form. (All one thousand elements except 5).
Case 2. Let k = 2, thus {5k, a} = {10, a}. Again, there are 999 subsets of this form. However, the subset {10, 5} is already counted in the first case. Thus we only count the subsets of this form that are not in the previous case. There are 998 such subsets. (All one thousand elements except 5 and 10).
.. .
Case 200. Let k = 200 thus {5k, a} = {1000, a}. There are 800 subsets that are distinct from the previous cases. (All one thousand elements except 5, 10, 15, . . . , 990, 995 and 1000).
Thus there are 999 + 998 + 997 + · · · + 801 + 800 = 179900 2-element subsets of X whose product of the elements is divisible by 5.
TT We shall partition the set X into two sets: X1= {5, 10, 15, . . . , 995, 1000}and X2= {1, 2, 3, 4, 6,
. . . , 997, 998, 999}. Note that all elements in X1 are divisible by 5 while those in X2are not.
Fur-thermore, |X1| = 200 and |X2| = 800. To get two element subsets of X such that the product its
elements is divisible by 5, we either get two elements from set X1, which can be done in 2002
ways, or get one element each from set X1and X2, which can be done in 2001
800
1 ways. Thus
by AP, the total number of such subsets is 2002 + 2001 8001 = 179900. 33. Consider the following set of points in the xy-plane:
A = {(a, b) | a, b ∈ Z and |a| + |b| ≤ 2}. Find
(i) |A|;
(ii) the number of straight lines which pass through at least 2 points in A; and (iii) the number of triangles whose vertices are points in A.
34. Let P be a convex n-gon, where n ≥ 6. Find the number of triangles formed by any 3 vertices of P that are pairwise nonadjacent in P .
35. 6 boys and 5 girls are to be seated around a table. Find the number of ways that this can be done in each of the following cases:
(i) There are no restrictions; PB (11 − 1)! = 10! ways. (ii) No 2 girls are adjacent;
PB We first arrange the 6 boys around the table, then insert the girls in between the boys. This guarantees that no two girls will be seated next to each other. Hence we have: 5! 655! ways.
(iii) All girls form a single block;
PB We first chunk all the girls and treat them as one “object”. We now have a total of 7 “objects”, which we have to arrange around the table. We then permute the girls inside the chunk. Hence we have: 6!5! ways.
(iv) A particular girl G is adjacent to two particular boys B1and B2.
PB We first chunk [B1GB2], so now we have a total of 9 “objects”, which we have to arrange
around the table. Then we permute B1and B2. Hence we have: 8!2! ways.
36. Show that the number of r-circular permutations of n distinct objects, where 1 ≤ r ≤ n, is given by n!
(n−r)!·r.
37. Let k, n ∈ N. Show that the number of ways to seat kn people around k distinct tables such that there are n people in each table is given by (kn)!nk .
38. Let r ∈ N such that
1 9 r − 1 10 r = 11 6 11r . Find the value of r.
FO 1 9! r!(9−r)! − 10!1 r!(10−r)! = 11 6( 10! r!(11−n)!) ⇐⇒ r!(9 − r)! 9! − r!(10 − r)! 10! = (11)(r!)(11 − r)! 6(11!) ⇐⇒ r!(9 − r)! 9! 1 −(10 − r) 10 = (11)(r!)(11 − r)! 6(11!) ⇐⇒ 9! r!(9 − r)! r!(9 − r)! 9! r 10 = (11)(r!)(11 − r)(10 − r)(9 − r)! 6(11)(10)(9!) 9!(10) r!(9 − r)! ⇐⇒ r = (11 − r)(10 − r) 6 ⇐⇒ 6r = 110 − 21r + r 2⇐⇒ r2− 27r + 110 = 0 ⇐⇒ (r − 22)(r − 5) = 0 ⇐⇒ r = 22or r = 5
But it is impossible for nr where r > n — by definition, r ≤ n. Thus, r = 5.
39. Prove each of the following identities: (a) n r = n r n − 1 r − 1 , where n ≥ r ≥ 1; LL n r n − 1 r − 1 =n r (n − 1)! (r − 1)!(n − 1 − r + 1)! = n! r!(n − r)! = n r .
(b) n r = n − r + 1 r n r − 1 , where n ≥ r ≥ 1; LL n − r + 1 r n r − 1 = n − r + 1 r n! (r − 1)!(n − r + 1)! = n!(n − r + 1) r!(n − r + 1)! = n! r!(n − r)! = n r . (c) n r = n n − r n − 1 r , where n > r ≥ 0; LL n n − r n − 1 r = n n − r (n − 1)! r!(n − 1 − r)! = n! r!(n − r)! = n r . (d) n m m r =n r n − r m − r , where n ≥ m ≥ r ≥ 0. LL n r n − r m − r = n! r!(n − r)! (n − r)! (m − r)!(n − r − m + r)! = n!(n − r)! r!(n − r)!(m − r)!(n − m)!· m! m! = n! m!(n − m)!· m! r!(m − r)!= n m m r .
40. Prove the identity nr = n
n−r by using the bijection principle.
41. Let X = {1, 2, . . . , n}, A = {A ⊂ X | n /∈ A}, and B = {A ⊂ X | n ∈ A}. Show that |A| = |B| by using the bijection principle.
MS For every A ∈ A, X \ A ∈ B; for every B ∈ B, X \ B ∈ A. Therefore, |A| = |B|. 42. Let r, n ∈ N. Show that the product
(n + 1)(n + 2) · · · (n + r) of r consecutive positive integers is divisible by r!.
MS (n + 1)(n + 2) · · · (n + r) = Pn+r r = r!
n+r
r . Hence, r! | (n + 1)(n + 2) · · · (n + r).
43. Let A be a set of kn elements, where k, n ∈ N. A k-grouping of A is a partition of A into k-element subsets. Find the number of different k-groupings of A.
KM For the first k-subset, knk. For the 2nd k-subset, kn−k
k . For the 3rd k-subset, kn−2k
k . And
for the nth k-subset, kn−(n−1)kk = k
k. Hence, the number of different k-groupings would seem
to be knk kn−k k
kn−2k k · · ·
k
k. But we still need to divide by the number of ways to arrange the
n k-subsets since the the ordering of the k-subsets doesn’t matter. Therefore, the final answer is
kn k kn−k k kn−2k k · · · k k n! .
44. Twenty-five of King Arthur’s knights are seated at their customary round table. Three of them are chosen — all choices of three being equally likely — and are sent off to slay a troublesome dragon. Let P be the probability that at least two of the three had been sitting next to each other. If P is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
KM Number in sample space = 253 = 2300. How many ways can we choose 3 knights such that at least two of the three had been sitting next to each other (event E)?
Case 1: There are exactly 2 knights sitting next to each other. There are 25 choices for the first knight, there are 2 ways to choose a seatmate (his left or right), and there are 21 ways to select another knight which is not adjacent to either of the knights already chosen. Divide by 2! since we counted twice the possibility that the seatmate was chosen first and the original was chosen next. Ans: (25)(2)(21)2! = 525.
Case 2: There are exaclty 3 knights sitting next to each other. Ans: There are 25 such consecutive ”triples”.
Therefore, P (E) = 525 + 25 2300 =
550 2300 =
11
46 and the sum of the numerator and the denominator is 57.
45. One commercially available ten-button lock may be opened by depressing — in any order — the correct five buttons. The sample shown below has {1, 2, 3, 6, 9} as its correct combination. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?
1 2 3 4 5 6 7 8 9 10 a a a a a a a a a a a a a a a
MS Currently, the locks require five buttons to unlock, which indicates that there are 105 com-binations. If, after redesigning, as few as one button to as many as nine buttons can be used as a combination, only two combinations will not be viable: the combination where no button is depressed 100, and the combinations where all buttons are depressed 10
10. Hence, the number
of combinations added by the redesign is 101 + 10 2 + 10 3 + 10 4 + 10 6 + 10 7 + 10 8 + 10 9 = 770.
46. In a shooting match, eight clay targets are arranged in two hanging columns of three each and one column of two, as pictured. A marksman is to break all eight targets accoring to the following rules: (1) The marksman first chooses a column from which a target is to be broken. (2) The marksman must then break the lowest remaining unbroken target in the chosen column. If these rules are followed, in how many different orders can the eight targets be broken?
i i i
i i i
i i
TT Consider a string of letters that will be composed of three As, two Bs and three Cs. The number of ways in which we can construct such a string is equal to 3!·2!·3!8! .
Now let’s note a particular string, say ABACCABC. If we are to relate this to the given problem, we can treat the three As as the clay targets on the first column, the two Bs as the clay targets on the second column, and the three Cs as the clay targets on the third column.
Considering the given arrangement of letters in the string, this can be interpreted as: (1) on the first shot, the first (or bottom) clay target on the first column was hit; (2) on the second shot, the first (or bottom) clay target of the second column was hit; (3) on the third shot, the second (or middle) clay target of the first column was hit; (4) on the fourth shot, the first (or bottom) clay target of the third column was hit; (5) on the fifth shot, the second (or middle) clay target of the third column was hit; (6) on the sixth shot, the third (or top) clay target of the first column was hit;
(7) on the seventh shot, the second (or top) clay target of the second column was hit; (8) and finally on the last shot, the remaining clay target on the third column was hit. Note that the interpretation is still consistent with the rules stated in the problem.
Therefore, we can say that there is a bijection between the number of ways in which we can arrange three As, two Bs and three Cs to form a string, and the number of ways in which a shooter can break all the 8 clay targets according to the rules.
Thus the number of ways a shooter can accomplish the task is equal to 3!·2!·3!8! = 560.
47. Using the numbers 1, 2, 3, 4, 5 we can form 5!(= 120) 5-digit numbers in which the 5 digits are all distinct. If these numbers are listed in increasing order:
12345
1st , 123542nd , 124353rd , . . . , 54321120th,
find (i) the position of the number 35421; (ii) the 100th number in the list. FO 12345 12543 13524 14352 15324 21354 23145 12354 13245 13542 14523 15342 21435 23154 12435 13254 14235 14532 15423 21453 23415 12453 13425 14253 15234 15432 21534 23451 12534 13452 14325 15243 21345 21543 23514 23541 31524 35241 43215 52341 24135 31542 35412 43251 52314 24153 32145 35421 43512 52413 24315 32154 41235 43521 52431 24351 32415 41253 45123 53124 24513 34251 41325 45132 53142 24531 32514 41352 45213 53214 25134 32541 41523 45231 53241 25143 34125 41532 45312 54123 25314 34152 42135 45321 54132 25341 34215 42153 51234 54213 25413 34251 42315 51243 54231 25431 34512 42351 51324 54312 31245 34521 42513 51342 54321 31254 35124 42531 51423 31425 35142 43125 52134 31452 35214 43152 52143
(i) The position of the number 35421 is in the 72nd position. (ii) the 100th number in the list is the number 51342.
48. The P4
3(= 24)3-permutations of the set {1, 2, 3, 4} can be arranged in the following way, called the
lexicographic ordering:
123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, . . . , 431, 432.
Thus the 3-permutations “132” and “214” appear at the 3rd and 8th positions of the ordering respectively. There are P9
4(= 3024)4-permutations of the set {1, 2, . . . , 9}. What are the positions
of the permutations “4567” and “5182” in the corresponding lexicographic ordering of the 4-permutations of the set {1, 2, . . . , 9}?
KM Let us consider first 4567. We will find all the permutations less than 4567. There are (3)(8)(7)(6) = 1008numbers less than 4000. (The thousands place can only come from {1, 2, 3}.) There are (1)(3)(7)(6) = 126 numbers greater than 4000 but less than 4500. (The hundreds can only be {1, 2, 3}. The thousands place is already 4.) There are (1)(1)(3)(6) = 18 numbers greater than 4500 but less than 4560. (The tens can only be {1, 2, 3}.) Lastly, there are (1)(1)(1)(3) = 3 numbers greater than 4560 but less than 4567. (The ones can come from {1, 2, 3}.) Thus, the total number of permutations less than 4567 totals to 1155, implying that 4567 is the 1156th permutation in lexicographic ordering.
Finding the position of 5182 is done similarly. There are (4)(8)(7)(6) = 1344 numbers less than 5000. (The thousands can be {1, 2, 3, 4}.) There are no permutations greater than 5000 but less than 5100. There are (1)(1)(5)(6) = 30 numbers greater than 5100 and less than 5180. (The tens can come from {2, 3, 4, 6, 7} and the ones can come from those remaining from the choices for the tens (4), along with 8 and 9. This gives 6 choices for the ones.) Lastly, there are no permutations greater than 5180 but less than 5182. Thus, the total number of permutations less than 5182 totals to 1374, implying that 5182 is the 1375th permutation in lexicographic ordering.
49. The 53(= 10) 3-element subsets of the set {1, 2, 3, 4, 5} can be arranged in the following way, called
the lexicographic ordering:
{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}.
Thus the subset {1, 3, 5} appears in the 5th position of the ordering. There are 104 4-element
subsets of the set {1, 2, . . . , 10}. What are the positions of the subsets {3, 4, 5, 6} and {3, 5, 7, 9} in the corresponding lexicographic ordering of the 4-element subsets of {1, 2, . . . , 10}?
PB Notice that there are six 3-element subsets of the form {1, a, b}, where a and b are distinct elements of the set {2, 3, 4, 5}. Since the subset {2, 3, 4} appears right next to these six subsets, then it should appear in the 7th position. Keeping this in mind, it would now be easy for us to determine the position of a given subset.
There are 93 4-element subsets of the form {1, a, b, c}, where a, b and c are distinct elements of the set {2, 3, . . . , 10}; and there are 83 4-element subsets of the form {2, d, e, f }, where d, e and f are distinct elements of the set {3, 4, . . . , 10}. Also, the subset {3, 4, 5, 6} is the first subset starting with 3. Hence the subset {3, 4, 5, 6} appears in the 9
3 + 8
3 + 1 = 141th position.
More so, there are 62 4-element subsets of the form {3, 4, g, h}, where g and h are distinct ele-ments of the set {5, 6, . . . , 10}; and 41 4-element subsets of the form {3, 5, 6, i}, where i is an ele-ment of the set {7, 8, 9, 10}. Since the subset {3, 5, 7, 9} is the second subset of the form {3, 5, 7, j}, where j is an element of the set {8, 9, 10}, then it should appear in the position numbered
9 3 +8 3 +6 2 +4 1 + 2 = 161.
50. Six scientists are working on a secret project, They wish to lock up the documents in a cabinet so that the cabinet can be opened when and only when three or more of the scientists are present. What is the smallest number of locks needed? What is the smallest number of keys each scientist must carry?
51. A 10-storey building is to be painted with some 4 different colors such that each storey is painted with one color. It is not necessary that all the 4 colors must be used. How many ways are there to paint the building if
(i) there are no other restrictions?
PB Each floor has four choices, hence there are 410ways to paint the building.
(ii) any 2 adjacent stories must be painted with different colors?
PB Suppose we start painting with the ground floor, then initially we have four choices. But for the second floor, we only have three choices not including our previous choice of color. Again, for the third floor, we only have three choices, and so on. Thus there are 4 · 39ways to
paint the building.
52. Find the number of all multi-subset of M = {r1· a1, r2· a2, . . . , rn· an}.
TT To find the number of all multi-subsets of M = {r1· a1, r2· a2, . . . , rn· an}, note that, for every
object ai, there are (1 + ri)ways of “choosing” it — that is, the multisubset of M may contain none
of object ai, only one of object ai, two of object ai, up untill all of object ai(for a total (1 + ri)ways in
which object aican be a part of the multi-subset). Thus by (MP), the total number of multi-subsets
of M is equal to (1 + r1)(1 + r2) · · · (1 + rn).
53. Let r, n ∈ N with r ≤ n. A permutation x1x2· · · x2n of the set {1, 2, . . . , 2n} is said to have the
property P (r) if |xi− xi+1| = r for at least one i in {1, 2, . . . , 2n − 1}. Show that, for each n and r,
there are more permutations with property P (r) than without.
54. Prove by combinatorial argument that each of the following numbers us always an integer for each n ∈ N:
that k are of type 1, k are of type 2, . . . , k are of type n. And since these expressions are a result of counting the number of arrangements, then they must always be an integer.
Note that the number of ways to arrange such nk objects is given by (kn)! k! · k! · · · k! | {z } ntimes . (i) (3n)! 2n· 3n, TT (3n)! 2n· 3n = (3n)! (3!)n = (3n)! 3! · 3! · · · 3! | {z } ntimes .
This can viewed as the number of ways of arranging 3n objects, such that 3 are of type 1, 3 are of type 2, . . . , 3 are of type n.
(ii) (6n)! 5n· 32n· 24n, TT (6n)! 5n· 32n· 24n = (6n)! (6!)n = (6n)! 6! · 6! · · · 6! | {z } ntimes .
This can viewed as the number of ways of arranging 6n objects, such that 6 are of type 1, 6 are of type 2, . . . , 6 are of type n.
(iii) (n 2)! (n!)n, TT (n2)! (n!)n = (n2)! n! · n! · · · n! | {z } ntimes .
This can viewed as the number of ways of arranging n2objects, such that n are of type 1, n are of type 2, . . . , n are of type n. Note that the total number of objects, given you have n kinds, each with n objects, is n · n = n2.
(iv) (n!)! (n!)(n−1)!. TT (n!)! (n!)(n−1)! = [n · (n − 1)!]! n! · n! · · · n! | {z } (n−1)!times .
This can viewed as the number of ways of arranging n! objects, such that n are of type 1, n are of type 2, . . . , n are of type (n − 1)!. Note that the total number of objects, given you have (n − 1)!kinds, each with n objects, is n · (n − 1)! = n!.
55. Find the number of r-element multi-subsets of the multi-set
M = {1 · a1, ∞ · a2, ∞ · a3, . . . , ∞ · an}.
TT An r-element multi-subset of the multi-set M = {1 · a1, ∞ · a2, . . . , ∞ · an} may either (1) contain
a1, or (2) not contain a1.
Let xi represent the number of object ai in the subset. To determine the number of subsets
in (1), we only need to consider the number of non-negative integer solutions to the equation x1+ x2+ · · · + xn= r, where x1= 1. This is just equal to (r−1)+(n−1)−1r−1 = r+n−3r−1 . To determine
the number of subsets in (2), we determine the number of non-negative integer solutions to the equation x2+ · · · + xn= r. This is just equal to (r)+(n−1)−1r = r+n−2r .
Hence by (AP), the total number of r-element multi-subsets of M is equal to r+n−3r−1 + r+n−2 r .
56. Six distinct symbols are transmitted through a communication channel. A total of 18 blanks are to be inserted between the symbols with at least 2 blanks between every pair of symbols. In how many ways can the symbols and the blanks be arranged?
PB We first permute the six distinct symbols then fix the ten blanks in such a way that exactly two blanks are between every pair of symbols. This leaves us with eight more blanks to be inserted in five spaces. Therefore the number of ways to arrange the symbols and the blanks is given by:
8+5−1
5−1 6! = 356400.
57. In how many ways can the following 11 letters: A, B, C, D, E, F, X, X, X, Y, Y be arranged in a row so that every Y lies between two Xs (not necessarily adjacent)?
PB We first fix the three Xs. Since we want every Y to be in between two Xs, then we insert the Ys into the spaces between the Xs. Next, we insert the remaining six letters anywhere between the Xs and Ys including the “outside” spaces (space to the left of the leftmost X and to the right of the rightmost X). Finally, we permute these six letters. Hence we have: 2+2−12 6+6−1
6 6!.
58. Two n-digit integers (leading zero allowed) are said to be equivalent if one is a permutation of the other. For instance, 10075, 01057 and 00751 are equivalent 5-digit integers.
(i) Find the number of 5-digit integers such that no two are equivalent.
TT The number of 5-digit integers such that no two are equivalent is just the same as the number of 5-element multi-subsets of M = {∞ · 0, ∞ · 1, . . . , ∞ · 9}, or the number of non-negative integer solutions to the equation x0+ x1+ · · · + x9= 5, where xidenotes the number
of times digit i was used, which is just equal to 145.
(ii) If the digits 5, 7, 9 can appear at most once, how many nonequivalent 5-digit integers are there?
TT If 5, 7, 9 can appear at most once, we shall consider four cases: (a) none of 5, 7, 9 are included
The number of such 5-digit integers is equal to the number of non-negative integer solu-tions to the equationP9
i=0 i6=5,7,9
xi= 5. This will give us 115 numbers.
(b) one of the numbers 5, 7, 9 is included
The number of such 5-digit integers is equal to the number of non-negative integer solu-tions to the equationP9
i=0 i6=5,7,9
xi
+ xs = 5, where xs= 1, for s = 5, 7 or 9. This will give
us 31 10
4 numbers.
(c) two of the numbers 5, 7, 9 are included
The number of such 5-digit numbers is equal to the number of non-negative integer solu-tions to the equationP9
i=0 i6=5,7,9
xi
+ (xs+ xt) = 5, where xs= xt= 1, for s, t = 5, 7, 9 and
s > t. This will give us 32 9
3 numbers.
(d) all of the numbers 5, 7, 9 are included
The number of such 5-digit numbers is equal to the number of non-negative integer solu-tions to the equationP9
i=0 i6=5,7,9
xi= 2. This will give us 82 numbers.
Hence, the total number of such 5-digit numbers is equal to 115 + 3 10 4 + 3
9 3 +
8 2.
59. How many 10-letter words are there using the letters a, b, c, d, e, f if (i) there are no restrictions?
KM, LL 10-letter word, each letter has 6 choices. Ans: 610.
(ii) each vowel (a and e) appears 3 times and each consonant appears once? KM Just count all possible arrangements of the word aaaeeebcdf. Ans: 10!
3!3!.
LL The number of such 10-letter words is the number of permutations of the multi-set {3 · a, b, c, d, 3 · e, f}, which is equal to 3!3!10!.
(iii) the letters in the word appear in alphabetical order?
KM This is bijective to looking for all nonnegative integer solutions to x1+ x2+ · · · + x6= 10.
Ans: H6 10= 10 + 6 − 1 10 =15 10 .
(iv) each letter occurs at least once and the letters in the word appear in alphabetical order? KM This is bijective to looking for all the positive integer solutions to x1+ x2+ · · · + x6= 10.
Ans: H6 4= 4 + 6 − 1 4 =9 4 .
60. Let r, n, k ∈ N such that r ≥ nk. Find the number of ways of distributing r identical objects into n distinct boxes so that each box holds at least k objects.
FO k 1 k 2 k 3 · · · k n
First put k objects in each box to fulfill the requirements. This can be done in one way. Then distribute the remaining r − nk objects in the boxes in an arbitrary way. Using (MP) and result in case 2(ii) r − nk + n − 1 r − nk =r − n(k − 1) − 1 r − nk ; by identity (1.4.2) this can also be written as
r − n(k − 1) − 1 n − 1
.
61. Find the number of ways of arranging the 9 letters r, s, t, u, v, w, x, y, z in a row so that y always lies between x and z (x and y, or y and z need not be adjacent in the row).
PB We first fix x, y and z respectively so that y is always between x and z. We then permute xand z. Next we insert the remaining six letters to the spaces between x, y and z, including the “outside” spaces. Finally we permute these six letters. Hence we have: 2! 4+6−16 6!.
62. Three girls A, B and C, and nine boys are to be lined up in a row. In how many ways can this be done if B must lie between A and C, and A, B must be separated by exactly 4 boys?
TT Since B is in between A and C, then the girls can only be arranged in two ways: CBA and ABC. Now consider a particular arrangement, ABC.
9 boys must be placed in between/beside the girls; assume for now that every boy is the same. Since there should be exactly 4 boys in between A and B, we are going to have AY Y Y Y BC (Y represents a boy). Next we’re going to place the boys to the left of A, between B and C, and to the right of C. There are 3 spaces where the 5 remaining boys can position themselves in, and the number of ways in which they can to that is given by 5+3−15 . Since the 9 boys are different from one another, by MP we shall have 5+3−15 · 11!.
Since the number of ways of arranging the boys satisfying the given conditions given the initial arrangement of girls ABC is the same as that for CBA, then the total number of arrangements is just given by 5+3−15 · 11! · 2 = 42 · 11!.
63. Five girls and eleven boys are to be lined up in a row such that, from left to right, the girls are in the order G1, G2, G3, G4, G5. In how many ways can this be done if G1and G2must be separated
by at least 3 boys, and there is at most one boy between G4and G5?
64. Given r, n ∈ N with r ≥ n, let L(r, n) denote the number of ways of distributing r distinct objects into n identical boxes so that no box is empty and the objects in each box are arranged in a row. Find L(r, n) in terms of r and n.
KM The number of ways to place r identical objects into n distinct boxes is r−1n−1. Multiply by r! since the objects must be distinct and the arrangement matters. Divide by n! since the boxes must be identical and hence, its arrangements must be removed from the counting. Therefore, L(r, n) = n!r! n−1r−1.
65. Find the number of integer solutions to the equation:
x1+ x2+ x3+ x4+ x5+ x6= 60
in each of the following cases:
(i) xi≥ i − 1 for each i = 1, 2, . . . , 6;
PB By subtracting i − 1 to xifor each i = 1, 2, . . . , 6, we have adjusted the lower bound for
each xito zero instead of i − 1. However, for the original equation to remain constant, we also
have to subtractP6
i=1(i − 1)from 60. Hence the number of integer solutions to the equation
given above is the same to the number of nonnegative integer solutions to the equation: y1+ y2+ y3+ y4+ y5+ y6= 45.
Hence we have 45+6−145 = 50 5.
(ii) x1≥ 2, x2≥ 5, 2 ≤ x3≤ 7, x4≥ 1, x5≥ 3 and x6≥ 2.
PB Again we have to adjust each lower bound to zero instead of some other number for our general result to work. However, notice that x3has an upper bound. Let us ignore this first
for the meantime. By the same technique shown above, the number of integer solutions to the equation above (ignoring x3’s upper bound) is the same as the number of nonnegative
integer solutions to the equation:
y1+ y2+ y3+ y4+ y5+ y6= 60 − 2 − 5 − 2 − 1 − 3 − 2 = 45.
Hence we have 505. However, we still have to remove all the cases when x3 ≥ 8 and there
are 39+6−139 = 44
5 of them. Hence our final answer is 50
5 − 44
5.
66. Find the number of integer solutions to the equation: x1+ x2+ x3+ x4= 30
in each of the following cases: (i) xi≥ 0 for each i = 1, 2, 3, 4;
PB From our general result, the number of nonnegative integer solutions to the equation above is given by: 30+4−130 = 33
3.
(ii) 2 ≤ x1≤ 7 and xi≥ 0 for each i = 2, 3, 4;
PB By the same technique shown in the previous number, we have: 28+4−128 − 22+4−1 22 = 31 3 − 25 3. (iii) x1≥ −5, x2≥ −1, x3≥ 1 and x4≥ 2.
PB Again, the number of integer solutions to this equation is the same as the number of nonnegative integer solutions to the equation:
y1+ y2+ y3+ y4= 30 + 5 + 1 − 1 − 2 = 33.
Hence we have: 33+4−133 = 36 3.
67. Find the number of quadruples (w, x, y, z) of nonnegative integers which satisfy the inequality w + x + y + z ≤ 1992.
TT, JA The given inequality can be converted to an equation by introducing a slack variable s, where s is nonnegative:
w + x + y + z + s = 1992.
The slack variable s captures any excess value not taken up by the original variables. That means when w + x + y + z = 1991, then s = 1; when w + x + y + z = 1987, then s = 5.
Since s is non-negative, it would not be possible for w + x + y + z > 1992, as this will force s to become negative. Thus the number of quadruplets (w, x, y, z) satisfying the inequality is the same the number of nonnegative integer solutions for the equation w + x + y + z + s = 1992. This can be interpreted as the number of ways of arranging 1992 sticks and 4 “+” signs, is equal to 19964 .
68. Find the number of nonnegative integer solutions to the equation: 5x1+ x2+ x3+ x4= 14.
JA, LL Consider the following cases:
Case 1: What if x1= 0? Then the equation becomes
x2+ x3+ x4= 14.
In this case, we have
n + r − 1 r − 1 =14 + 2 2 =16 2 . Case 2: What if x1= 1? Then the equation becomes
5(1) + x2+ x3+ x4= 14,
which is equal to
x2+ x3+ x4= 9,
and by the same reasoning, we have n + r − 1 r − 1 =9 + 2 2 =11 2 . Case 3: What if x1= 2? Then the equation becomes
5(2) + x2+ x3+ x4= 14 ⇔ x2+ x3+ x4= 4.
In this case, we have
n + r − 1 r − 1 =4 + 2 2 =6 2 . So, by A.P., we have,
16 2 +11 2 +6 2 = 120 + 55 + 15 = 190. 69. Find the number of nonnegative integer solutions to the equation:
rx1+ x2+ · · · + xn= kr,
where k, r, n ∈ N.
JA, LL Consider the following cases:
Case 1: What if x1= 0? Then the equation becomes
x2+ · · · + xn= kr.
In this case, we have n − 2 pluses and so
kr + n − 2 n − 2
. Case 2: What if x1= 1? Then the equation becomes
r(1) + x2+ · · · + xn= kr
which is equal to
x2+ · · · + xn = kr − r = r(k − 1)
and by the same reasoning, we have
r(k − 1) + n − 2 n − 2
.
We do this inductively until we reach this case: Case k + 1: What if x1= k? Then the equation becomes
r(k) + x2+ · · · + xn= kr
which is equal to
x2+ · · · + xn= kr − kr = r(k − k)
and by the same reasoning, we have
r(k − k) + n − 2 n − 2 =n − 2 n − 2 .
Getting a general form of this (after some thought) we have,
k X i=0 r(k − i) + n − 2 n − 2 .
70. Find the number of nonnegative integer solutions to the equation: 3x1+ 5x2+ x3+ x4= 10.
PB Since each xi ≥ 0 for i = 1, 2, 3, 4, then 0 ≤ x1≤ 3 and 0 ≤ x2≤ 2. We now consider 4 · 3 = 12
cases:
(x1, x2) # of nonnegative integer solutions (x1, x2) # of nonnegative integer solutions
(0, 0) 10+2−110 = 111 = 11 (1, 1) 2+2−12 = 31 = 3 (1, 0) 7+2−17 = 8 1 = 8 (1, 2) ∅ → 0 (2, 0) 4+2−14 = 5 1 = 5 (2, 1) ∅ → 0 (3, 0) 1+2−11 = 2 1 = 2 (2, 2) ∅ → 0 (0, 1) 5+2−15 = 61 = 6 (3, 1) ∅ → 0 (0, 2) 0+2−10 = 1 1 = 1 (3, 2) ∅ → 0
Hence there are 36 nonnegative integer solutions to the equation above. 71. Find the number of positive integer solutions to the equation:
(x1+ x2+ x3)(y1+ y2+ y3+ y4) = 77.
JA Consider the following cases:
Case 1: If x1+ x2+ x3= 7and y1+ y2+ y3+ y4= 11, then the number of nonnegative integer solutions
to the equation x1+ x2+ x3= 7is given by
r + n − 1 r =7 + 3 − 1 7 =9 7 =9 2 , and to the equation y1+ y2+ y3+ y4= 11is given by
r + n − 1 r =11 + 4 − 1 11 =14 11 =14 3 . By M.P., the desired number is given by
9 2 14 3 .
Case 2: If x1+ x2+ x3= 11and y1+ y2+ y3+ y4= 7, then the number of nonnegative integer solutions
to the equation y1+ y2+ y3+ y4= 7is given by
r + n − 1 r =7 + 4 − 1 7 =10 7 =10 3 , and to the equation x1+ x2+ x3= 11is given by
r + n − 1 r =11 + 3 − 1 11 =13 11 =13 2 . By M.P., the desired number is given by
10 3 13 2 .
By A.P., the desired number of nonnegative integer solutions to the equation (x1+ x2+ x3)(y1+ y2+ y3) = 77 is given by 10 3 13 2 +9 2 14 3 . 72. Find the number of nonnegative integer solutions to the equation:
(x1+ x2+ · · · + xn)(y1+ y2+ · · · + yn) = p,
where n ∈ N and p is a prime.
PB Since p is a prime, then only either x1+ x2+ · · · + xn = pand y1+ y2+ · · · + yn = 1or
x1+ x2+ · · · + xn= 1and y1+ y2+ · · · + yn= p. Hence we have:
p + n − 1 p 1 + n − 1 1 +1 + n − 1 1 p + n − 1 p = 2np + n − 1 p .
73. There are 5 ways to express “4” as a sum of 2 nonnegative integers in which the order counts: 4 = 4 + 0 = 3 + 1 = 2 + 2 = 1 + 3 = 0 + 4.
Given r, n ∈ N, what is the number of ways to express r as a sum of n nonnegative integers in which the order counts?
74. There are 6 ways to express “5” as a sum of 3 positive integers in which the order counts: 5 = 3 + 1 + 1 = 2 + 2 + 1 = 2 + 1 + 2 = 1 + 3 + 1 = 1 + 2 + 2 = 1 + 1 + 3.
Given r, n ∈ N with r ≥ n, what is the number of ways to express r as a sum of n positive integers in which the order counts?
75. A positive integer d is said to be ascending if in its decimal representation: d = dmdm−1· · · d2d1, we
have
0 < dm≤ dm−1≤ · · · ≤ d2≤ d1.
For instance, 1337 and 2455566799 are ascending integers. Find the number of ascending integers which are less than 109.
TT For this problem, we can only have ascending integers with at most nine digits. To determine all such integers, let us consider a seqeunce of 9 sticks and 9 “+” signs.
If we were to initially place all nine “+” signs in a row, there will be 10 spaces available for the sticks to occupy (8 spaces in between the “+” signs and 2 spaces at the ends of the row). Relating this to the given problem, we can actually view the first 9 spaces as the numbers from 1 to 9 (i.e., the
first space represents the digit 1, the 2nd space represents the digit 2, and so on). As for the 9 sticks, they determine the number of digits that the ascending number will have.
In particular, if we let 7 of the sticks occupy (in any manner) the first 9 spaces — and therefore the 2 remaining sticks occupy the last space — this means that we’re actually forming a 7-digit number. If only 1 stick occupies any of the first 9 spaces — and therefore the 8 remaining sticks will occupy the last space — this means that we’re just forming a 1-digit number. Notice that the more sticks placed in the last space, the lesser the number of digits that the number will have. Moreover, we can only construct a number with at most 9 digits, since we only have nine sticks. To illustrate this, consider the sequence
|| + ||| + ||| + + + + + + + |.
Using the above interpretation, we can say that the number contains 8 digits, composed of two 1s, three 2s, three 3s. Note, however, that there is only one way in which we can construct an ascending number, and that is 11222333.
Here’s another sequence:
+ + | + +| + + + + + |||||||.
This can be interpreted as an ascending number containing only two digits, and it is equal to 35. Thus, the number of ways of arranging 9 sticks and 9 “+” signs can give us the total number of ascending numbers less than 109, which is 18
9. However, we have to eliminate the sequence
+ + + + + + + + +||||||||||, since it does not correspond to any ascending number.
Thus, the total number of ascending numbers less than 109is 18 9 − 1.
76. A positive integer d is said to be strictly ascending if in its decimal representation: d = dmdm−1· · · d2d1, we have
0 < dm< dm−1< · · · < d2< d1.
For instance, 145 and 23689 are strictly ascending integers. Find the number of strictly ascending integers which are less than (i) 109, (ii) 105.
77. Let A = {1, 2, . . . , n}, where n ∈ N.
(i) Given k ∈ A, show that the number of subsets of A in which k is the maximum number is given by 2k−1.
(ii) Apply (i) to show that
n−1
X
i=0
2i = 2n− 1.
78. In a given circle, n ≥ 2 arbitrary chords are drawn such that no three are concurrent within the interior of the circle. Suppose m is the number of points of intersection of the chords within the interior. Find, in terms of n and m, the number r of line segments obtained through dividing the chords by their points of intersection.
79. There are p ≥ 6 points given on the circumference of a circle, and every two of the points are joined by a chord.
(i) Find the number of such chords.
Assume that no 3 chords are concurrent within the interior of the circle.
(ii) Find the number of points of intersection of these chords within the interior of the circle. (iii) Find the number of line segments obtained through dividing the chords by their points of
(iv) Find the number of triangles whose vertices are the points of intersection of the chords within the interior of the circle.
80. In how many ways can n + 1 different prizes be awarded to n students in such a way that each student has at least one prize?
FO Since there are n students, then they can be arranged n! ways for the prizes to be awarded. The fact that at least one prize each student then n+1r be number of distributions to be awarded at n students, where r = 2, for r > 2 there will be student with no prize at all. Thus n+12 n! by (MP) are the total number of ways.
81. (a) Let n, m, k ∈ N, and let Nk = {1, 2, . . . , k}. Find
(i) the number of mappings from Nnto Nm.
(ii) the number of 1-1 mappings from Nnto Nm, where n ≤ m.
(b) A mapping f : Nn → Nmis strictly increasing if f (a) < f (b) whenever a < b in Nn. Find the
number of strictly increasing mappings from Nnto Nm, where n ≤ m.
(c) Express the number of mappings from Nn onto Nmin terms of S(n, m) (the Stirling number
of the second kind).
82. Given r, n ∈ Z with 0 ≤ n ≤ r, the Stirling number s(r, n) of the first kind is defined as the number of ways to arrange r distinct objects around n identical circles such that each circle has at least one object. Show that
(i) s(r, 1) = (r − 1)! for r ≥ 1; (ii) s(r, 2) = (r − 1)!(1 +1 2+ 1 3+ · · · + 1 r−1)for r ≥ 2; (iii) s(r, r − 1) = r2 for r ≥ 2; (iv) s(r, r − 2) = 241r(r − 1)(r − 2)(3r − 1)for r ≥ 2; (v) Pr n=0s(r, n) = r!.
83. The Stirling numbers of the first kind occur as the coefficients of xnin the expansion of
x(x + 1)(x + 2) · · · (x + r − 1). For instance, when r = 3,
x(x + 1)(x + 2) = 2x + 3x2+ x3= s(3, 1)x + s(3, 2)x2+ s(3, 3)x3; and when r = 5, x(x + 1)(x + 2)(x + 3)(x + 4) = 24x + 50x2+ 35x3+ 10x4+ x5 = s(5, 1)x + s(5, 2)x2+ s(5, 3)x3+ s(5, 4)x4+ s(5, 5)x5. Show that x(x + 1)(x + 2) · · · (x + r − 1) = r X n=0 s(r, n)xn, where r ∈ N.
84. Given r, n ∈ Z with 0 ≤ n ≤ r, the Stirling number S(r, n) of the second kind is defined as the number of ways of distributing r distinct objects into n identical boxes such that no box is empty. Show that (i) S(r, 2) = 2r−1− 1; (ii) S(r, 3) = 1 2(3 r−1+ 1) − 2r−1; (iii) S(r, r − 1) = r2; (iv) S(r, r − 2) = r3 + 3 r 4.
85. Let (x)0= 1and for n ∈ N, let
(x)n= x(x − 1)(x − 2) · · · (x − n + 1).
The Stirling numbers of the second kind occur as the coefficients of (x)n when xris expressed in
terms of (x)ns. For instance, when r = 2, 3 and 4, we have, respectively,
x2 = x + x(x − 1) = (x)1+ (x)2 = S(2, 1)(x)1+ S(2, 2)(x)2 x3 = x + 3x(x − 1) + x(x − 1)(x − 2) = S(3, 1)(x)1+ S(3, 2)(x)2+ S(3, 3)(x)3 x4 = x + 7x(x − 1) + 6x(x − 1)(x − 2) + x(x − 1)(x − 2)(x − 3) = S(4, 1)(x)1+ S(4, 2)(x)2+ S(4, 3)(x)3+ S(4, 4)(x)4
Show that for r = 0, 1, 2, . . . ,
xr=
r
X
n=0
S(r, n)(x)n.
86. Suppose that m chords of a given circle are drawn in such a way that no three are concurrent in the interior of the circle. If n denotes the number of points of intersection of the chords within the circle, show that the number of regions divided by the chords in the circle is m + n + 1.
87. For n ≥ 4, let r(n) denote the number of interior regions of a convex n-gon divided by all its diagonals if no three diagonals are concurrent within the n-gon. For instance, as shown in the following diagrams, r(4) = 4 and r(5) = 11. Prove that r(n) = n4 +
n−1 2 . @ @ @ L L L LL ,, ,, , C C C C C CC l l l l l Z Z Z Z
88. Let n ∈ N. How many solutions are there in ordered positive integer pairs (x, y) to the equation xy
x + y = n?
PB Let n, x and y be elements of postive integers. Then the equation xy
x + y = n ⇐⇒ xy = n(x + y) ⇐⇒ xy − nx = ny ⇐⇒ x(y − n) = ny (88a) Thus, ny > 0 and so as x(y − n) > 0. This implies that y − n > 0, which enables us to write y in terms of n, i.e., y = n + b, for some positive integer b.
(88a) ⇐⇒ x = ny y − n, substituting n + a for y gives us:
x = n(n + b)
b ⇐⇒ x =
n2
b + n (88b)
However, x is a positive integer, then for (88b) to be valid, then b must divide n2, or in other words,
b must be a factor of n2. Thus this answers the problem: the number of solutions to the original
problem is the same as the number of positive factors or divisors of n2.
Furthermore, let n = p1k1· p2k2· · · pmkm, where pis are distint prime numbers and kis are positive
integers. Then n2would have (2k
1+ 1)(2k2+ 1) · · · (2km+ 1)factors. This is also the number of
89. Let S = {1, 2, 3, . . . , 1992}. In each of the following cases, find the number of 3-element subsets {a, b, c} of S satisfying the given condition:
(i) 3 | (a + b + c);
PB Express S = {1, 2, 3, . . . , 1992} in terms of its remainder when divided by 3, that is, we can write it in its “modulo-3” form, say S0 = {1, 2, 0, 1, 2, 0, . . . , 1, 2, 0}, where each number from S is represented in S0 by its remainder when divided by 3. Thus we have 664 elements of 1s, 2s and 0s in S0. We denote the set of all 1s by set A, all 2s by set B, and all 0s by set C. Therefore, to have the sum of any 3 elements from S0divisible by 3, we can simply choose all three numbers from the one set, i.e., all from either set A, set B or set C; or we choose one number from each of the three sets, i.e., one number from set A, one from set B and one from set C. Therefore, the number of ways to do this is given by
3664 3 +664 1 3 . (ii) 4 | (a + b + c).
PB Similarly, let S00 = {1, 2, 3, 0, 1, 2, 3, 0, . . . , 1, 2, 3, 0}, where each number from S is
repre-sented in S00 by its remainder when divided by 4. Thus we have 498 elements of 1s, 2s, 3s and 0s in S00. Again, we denote the set of all 1s by set A, all 2s by set B, all 3s by set C, and all 0s by set D. In order to have the sum of any three numbers from S00, we consider three cases. First, we can choose all three numbers from set D; second, we can choose exactly one number from set A, set C and set D, respectively; finally, we can choose two numbers from set C and one number from set B or two numbers from set A and one number from set B or two numbers from set B and one from set D. Therefore, the number of ways to do this is given by 498 3 +498 1 3 + 3498 1 498 2 .
90. A sequence of 15 random draws, one at a time with replacement is made from the set {A, B, C, . . . , X, Y, Z}
of the English alphabet. What is the probability that the string UNIVERSITY occurs as a block in the sequence?
PB The number of ways the string UNIVERSITY appears as a block is 265· 6 and the number
of ways to draw 15 letters with replacement is 2615. Thus the probability is 265· 6/2615 or simply
6/2610.
91. A set S = {a1, a2, . . . , ar} of positive integers, where r ∈ N and a1 < a2 < · · · < ar, is said to
be m-separated (m ∈ N) if ai− ai−1 ≥ m, for each i = 2, 3, . . . , r. Let X = {1, 2, . . . , n}. Find the
number of r-element subsets of X which are m-separated, where 0 ≤ r ≤ n − (m − 1)(r − 1). MS It is evident that there is only one 0-element subset of X, ∅, and there are n 1-element subsets of X, and all these subsets are vacuously m-separated. Let x1 = a1and xi = ai− ai−1
for 1 < i < r + 1 and xr+1= n − ar. Note that x1≥ 1, xr+1≥ 0, and for 1 < i < r + 1, xi≥ m and
x1+ x2+ · · · + xr+ xr+1 = n. There is a bijection between the collection of solutions S and the
collection of solutions x1, x2, . . . , xr+1; there is also a bijection between the collection of solutions
x1, x2, . . . , xr+1and the collection of solutions y1, y2, . . . , yr+1such that yi≥ 0 for 1 ≤ i ≤ r + 1 and
y1+ · · · + yr+1 = n − 1 − (r − 1)m, which has cardinality Hr+1n−1−(r−1)m = n−1−(r−1)m+r+1−1n−1−(r−1)m = n−(r−1)(m−1) r . Note: when r = 0, H 1 n+m−1= n+m−1 0 = 1; when r = 1, H 2 n−1= n 1 = n.
92. Let a1, a2, . . . , anbe positive real numbers, and let Skbe the sum of products of a1, a2, . . . , antaken
kat a time. Show that
SkSn−k≥
n k
2
