Combinatorial Probability
3.1 The Basic Counting Rule
Basic Exercises
3.1 a) The tree diagram is given by:
Model Elevation Outcome
•
•P
•A PA
•B PB
•C PC
•D PD
•V
•A VA
•B VB
•C VC
•D VD
•M
•A MA
•B MB
•C MC
•D MD
b) 12.
c) 3 × 4 = 12.
3.2 2 × 4 + 2 × 3 = 14.
3.3 5 × 3 × 6 × 4 = 360, by the BCR.
3.4 39 × 19 × 8 × 6 × 24 × 5 × 5 × 6 × 5 = 640, 224, 000.
3.5 a) 108, since there are 10 choices (i.e. 0, 1, · · · , 9) for each of the eight digits.
b) 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 1, 814, 400.
c) 10 × 97 = 47, 829, 690, since there are 10 choices for the 1st digit and 9 choices for every digit starting from the 2nd upto the 8th.
d) 10 × 9 × 86 = 23, 592, 960, since there are 10 choices for the 1st digit, 9 choices for the 2nd and 8 choices for each of the remaining six digits.
3.6 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 362, 880.
3.7 a) 104× 8 = 80, 000. b) 103× 9 = 9, 000. c) 107× 8 × 9 = 80, 000 × 9, 000 = 720, 000, 000.
3.8 a) 9×102= 900. b) 9×106 = 9, 000, 000. c) (9×102)×(9×106) = 81×108 = 8, 100, 000, 000.
3.9 a) 62 = 36.
b) 4 × 1 = 4, since, in order for the sum to be 5, the first die can have 1, 2, 3 or 4 dots facing up (i.e. a total of four choices) whereas the 2nd die must have exactly (5 − x1) dots facing up, where x1 is the number rolled on the 1st die.
c) 6.
d) 6 × 3 = 18. Note that the sum of the two dice is even if and only if either both dice show even numbers or both dice show odd numbers. Thus, although there are six possibilities for the 1st die, there are only three possibilities for the 2nd die (i.e. the 2nd die must show 1,3 or 5 if the 1st die shows an odd number, and the 2nd die must show 2,4 or 6 if the 1st die shows an even number).
3.10 a) 25 = 32. b) 5. c) 25− 1 = 31. d) 5+1=6.
3.11 a) 64 = 1296. b) 6 × 5 × 4 × 3 = 360. c) 2 × 63 = 432.
d) 64− 54 = 671, since there are 64 possible words when all six letters are used and there are 54 possible words when only five letters are used (i.e. when letter c is excluded).
3.12 a) nn, since each of the n elements in A can be mapped to n possible elements in B.
b) n × (n − 1) × · · · × 2 × 1. Since no two elements from A can be mapped to the same element in B, then while there are n possible choices for the value of the function at the 1st element in A, there are just (n − 1) possible choices for the value of the function at the 2nd element in A, thus there are only (n − 2) possible choices for the value of the function at the 3rd element in A, etc.
3.13 a) (5 × 4 × 3 × 2 × 1)/2 = 60. Note that the number of arrangements where c is before d is the same as the number of arrangements where c is behind d (due to symmetry). On the other hand, the total number of arrangements of five people in a line is 5 × 4 × 3 × 2 × 1.
b) 4 × 4 × 3 × 2 × 1 = 96.
3.14 The number of arrangements is equal to (n − k − 1) × 2 × ((n − 2)!) for k ∈ {0, 1, · · · , n − 2}, where (n − 2)! = (n − 2) × (n − 3) × · · · × 2 × 1. Indeed, among the two positions occupied by you and your best friend in the line, let x denote the position closest to the beginning of the line. Since there should be k people standing between you and your best friend, the possible values of x are 1, 2, · · · , (n − k − 1). Note that there are two choices regarding who (you or your
friend) is standing in position x. Finally, keeping the positions occupied by you and your friend fixed (i.e. positions numbered x, x + k + 1), there are (n − 2)! possibilities for arranging the remaining (n − 2) people in the (n − 2) available positions.
3.15 a) 4 × 3 × 2 × 1 × 24 = 384, since there are 4 × 3 × 2 × 1 ways to arrange the order of sitting for four married couples (where each couple is viewed as a single unit), and there are two sitting choices for each of the four couples (i.e. whether wife sits first and her husband sits next or the other way around).
b) (3 × 2 × 1) × 24 = 96. First let us view each married couple as a single unit. With respect to couple #1 sitting at the round table, there are 3 choices as to which couple (#2, #3 or
#4) is sitting to its right and then there are 2 choices as to which couple is sitting to its left (the remaining fourth couple has to sit in the position at the table opposite to couple #1). In addition, for every given arrangement of couples at the table, there are two choices per each couple as to whether a husband sits to the right from his wife or vice versa. The answer then follows by the BCR.
3.16 n +n(n−1)2 = n(n+1)2 , since there are n dominos which have equal numbers on both subrect-angles, and there are n(n−1) possible ordered pairs of distinct numbers selected from {1, · · · , n}, which results in n(n − 1)/2 possible dominos with unequal numbers on the two rectangles.
3.17 n(n − 1)/2. The number of possible handshakes is equal to the number of possible un-ordered pairs of people chosen from the n people at the party.
3.18 a) 52 × 51 × 50 × 49 × 48 = 311, 875, 200.
b) (52 × 51 × 50 × 49 × 48)/(5 × 4 × 3 × 2 × 1) = 2, 598, 960. The answer is obtained upon noting that (# of five-card draws) × (# of possible arrangements of five cards)= # of five-card studs.
3.19 a) n × (n − 1) × · · · × 1. Since none of the n boxes are left empty, then each box contains exactly one ball. Since balls and boxes are all distinguishable, then the problem is equivalent to finding the number of possible (ordered) arrangements of n (different) objects in a line.
b) n(n − 1)n(n − 1)
2 ((n − 2)!) = (n!)n(n − 1)
2 . Since exactly one box out of n is to be left empty, then one box contains 0 balls, one box contains two balls and (n − 2) boxes contain one ball each. There are n(n − 1) ways to choose the pair of boxes, where the first box is designated to contain 0 and the second box is designated to contain 2 balls, respectively. Next there are n(n − 1)/2 ways to choose the (unordered) set of two balls (from n available balls) that go into the box designated to contain two balls. Finally there are (n − 2)! = (n − 2)(n − 3) × · · · × 1 ways to arrange the remaining (n − 2) balls between the n − 2 boxes that contain one ball each.
The required answer then follows by the BCR.
c) nn− (n × (n − 1) × · · · × 1) = nn− (n!), since, in general, there are nn ways to assign n different balls to n different boxes (by Exercise 3.12(a)), while there are n! ways to arrange the balls so that no box is empty (by part (a) of the current problem).
Theory Exercises
3.20 To establish the BCR by method of mathematical induction, first note that the BCR was proved for r = 2 in Proposition 3.1. Next assume that the BCR holds for r = n, i.e. if there are m1 possibilities for the 1st action, and for each possibility of the 1st action there are
m2 possibilities for the 2nd action, and so on, then there are m = m1· m2· · · mn possibilities altogether for the n actions. Let us establish the BCR for r = n + 1. Let f1, · · · , fm (where m = m1· m2· · · mn) denote all possible outcomes occurring as a result of the first n actions. Let s1, · · · , smn+1 be the outcomes of the (n + 1)st action (taken by itself). Then we can enumerate the possibilities for all (n + 1) actions as follows:
f1s1 f1s2 . . . f1smn+1
f2s1 f2s2 . . . f2smn+1
... ... . .. ... fms1 fms2 . . . fmsmn+1,
where fksℓ represents selecting possibility fk, occurring as a result of the first n actions, and possibility sℓ for the (n + 1)st action. Each row in the display contains mn+1 possibilities and there are m rows, therefore, the total number of possibilities for the (n + 1) actions is equal to m · mn+1 = m1. . . mn· mn+1. The desired conclusion therefore follows.
Advanced Exercises
3.21 a) ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, thus, the total number of subsets of Ω = {a, b, c} equals 8.
b) 23 = 8. Each subset A ⊂ Ω is fully (and uniquely) determined by whether a ∈ A or a /∈ A, and by whether b ∈ A or b /∈ A, and by whether c ∈ A or c /∈ A. Since there are two possibilities for each of the three elements, then the required answer is 2 × 2 × 2.
c) 2n. Each subset A ⊂ Ω is fully (and uniquely) determined by whether each of the n elements of Ω belongs to A or not. Since there are two possibilities for each of the n elements, then the required answer is 2 × · · · × 2
! "# $
n
.
3.22 a) 2 × 8 = 24 = 16. For every subset A ⊂ {a, b, c}, one can construct two different subsets of {a, b, c, d}: A and A∪{d}. On the other hand, every subset of {a, b, c, d} can be constructed in that fashion from the subsets of {a, b, c}. Therefore, the number of possible subsets of {a, b, c, d}
is equal to twice the number of possible subsets of {a, b, c}.
b) Yes. Given an arbitrary subset A of E = {ω1, · · · , ωn}, one can construct two different subsets of {ω1, · · · , ωn, ωn+1}: A and A ∪ {ωn+1}. Moreover every subset of {ω1, · · · , ωn, ωn+1} can be constructed from subsets of E in that fashion. Therefore, the number of subsets of {ω1, · · · , ωn+1} is equal to twice the number of subsets of {ω1, · · · , ωn}.
c) For any set {ω1}, there are two possible subsets: ∅ and {ω1}. Thus, the number of subsets of a set, consisting of one element, is equal to 21 = 2. Now suppose that any set of n elements has 2nsubsets. Take an arbitrary set {ω1, · · · , ωn, ωn+1} of (n + 1) elements. Then the number of possible subsets of {ω1, · · · , ωn, ωn+1} is equal to 2 × (# of subsets of {ω1, · · · , ωn}) = 2 × 2n, by part (b). Thus, every set of n + 1 elements has 2n+1 subsets. The required conclusion then follows by induction.