Combinatorial Probability
3.2 Permutations and Combinations
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.
3.2 Permutations and Combinations
3.23 For all k, j ∈ N , with j < k,
k! = k(k − 1) · · · (k − j + 1) (k − j)(k − j − 1) · · · 2 · 1
! "# $
(k−j)!
= k(k − 1) · · · (k − j + 1)((k − j)!).
For k = j ∈ N , the following identity obviously holds: selections of the committee where the two specified people will have to serve together (there are 12 possibilities for selection of the third person in addition to the two specified people).
Another way of solving the problem is given by:
''2
represents the number of ways to select an unordered set of 3 people different from the two specified, %2
1
&%12
2
&
represents the number of ways to select an unordered set of 3 people of whom exactly one is one of the two specified. The sum has to be multiplied by 3! to obtain the number of ordered selections from the number of unordered selections.
3.35%6
ways to choose a set of three positions (out of six) that are occupied by letters. Once positions for letters (and, thus, for numbers) are fixed, there are 263 arrangements of letters in the three positions occupied by letters and there are 103 possible arrangements of numbers in the three positions occupied by numbers.
3.36 a) 10 × 4 = 40, since straight flush corresponds to poker hands of the following form:
{A, 2, 3, 4, 5}, {2, 3, 4, 5, 6}, · · · , {10, J, Q, K, A} (with five cards of the same suit), where A stands for ace, J for jack, Q for queen and K for king, thus one has ten different possibili-ties for each suit and there are four suits.
b) (13)2%4 denominations from the 13 available denominations and, for that pair of denominations, there are%4
4
&%4
1
&
ways to pick the suits of the cards.
c) (13)2%4
= 5108. (For each suit, take the number of possible hands of that suit and subtract the number of straight flushes of that suit).
e) 10(45− 4) = 10200.
3.37 Take the answers obtained in Exercise 3.36 and multiply by 5! to obtain the number of possible hands when the order in which the cards are received matters. Namely, one obtains:
Straight flushes (when order matters): 40(5!) = 4800;
Four of a kind (when order matters): 624(5!) = 74880;
Full house (when order matters): 3744(5!) = 449, 280;
Flush (when order matters): 5108(5!) = 612, 960;
Straight (when order matters): 10200(5!) = 1, 224, 000;
Three of a kind (when order matters): 54912(5!) = 6, 589, 440;
Two pair (when order matters): 123552(5!) = 14, 826, 240;
One pair (when order matters): 1098240(5!) = 131, 788, 800.
3.38 In Proposition 2.10, each sum of the form )
1≤k1<k2<···<kn≤NP (Ak1 ∩ Ak2 ∩ · · · ∩ Akn)
5! = 67, 800, 320. (Equivalently, one could compute %50
5
&
25, since one could pick the five states (out of 50), whose senators serve on the committee, and then pick one sen-ator out of two for each of the five states selected).
3.40 %3
3.41 a), b) Selection of k objects out of n without replacement is equivalent to dividing n objects into a “desired” and an “undesired” subcategories of sizes k and (n − k), respectively, which is in turn equivalent to selecting (n − k) objects (as “undesired”) out of n without re-placement. Therefore, %n
k
d) All three quantities in the equation provide expressions for computing the number of ways to choose j committee members who serve as (different) officers plus k −j committee members who are not officers, from a list of n candidates. Thus, the three expressions should be equal. Note
that the first one corresponds to first choosing all k committee members and then choosing j officers from among them. The second expression corresponds to first selecting k − j committee members who are not officers, and then selecting j people who will serve as committee officers from the remaining pool of n − (k − j) candidates. Finally, the third expression corresponds to first choosing j people who will serve as committee officers, and then choosing the remaining k − j committee members from the n − j candidates left.
e) Note that
'n k (
(k)j = n!
k!(n − k)! · k!
(k − j)! = n!
(k − j)!· 1
(n − k)! (3.1)
= n!
(k − j)!(n − (k − j))! ·(n − (k − j))!
(n − k)! =
' n
k − j (
(n − (k − j))j. Also
(n)j'n − j k − j (
= n!
(n − j)!· (n − j)!
(k − j)!(n − k)! = n!
(k − j)! · 1
(n − k)! ='n k (
(k)j, where the last equality follows from (3.1).
Theory Exercises
3.46 For arbitrary a1, · · · , an, b1, · · · , bn, we have that (a1+ b1)(a2+ b2) · · · (an+ bn) = *
k1,··· ,kn∈{0,1}
ak11. . . aknnb1−k1 1. . . b1−kn n
=
*n ℓ=0
*
k1,··· ,kn∈{0,1}:
k1+···+kn=ℓ
ak11. . . aknnb1−k1 1. . . b1−kn n.
Note that there are%n
ℓ
&
different ways to pick ℓ indices out of n and put corresponding kj’s equal to 1 (while the rest are zeros). Therefore, upon taking a1= · · · = an= a and b1 = · · · = bn= b, we obtain that
(a + b)n=
*n ℓ=0
*
k1,··· ,kn∈{0,1}:
k1+···+kn=ℓ
aℓbn−ℓ=
*n ℓ=0
'n ℓ (
aℓbn−ℓ.
3.47 a) Note that (a + b)0 = 1, (a + b)1 = a + b, (a + b)2 = a2+ 2ab + b2, (a + b)3= a3+ 3a2b + 3ab2+b3, (a+b)4 = a4+4a3b+6a2b2+4ab3+b4, (a+b)5 = a5+5a4b+10a3b2+10a2b3+5ab4+b5, (a + b)6 = a6+ 6a5b + 15a4b2+ 20a3b3+ 15a2b4+ 6ab5+ b6.
b) The Pascal’s triangle is given by:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
c) For all n, k ∈ N , with k ≤ n,
where the first equality holds since %1
j
&
= 0 for all j > 1, while the last equality holds by Exercise 3.47(c). Thus, the Vandermonde’s identity is verified for n = 1. Next assume that the Vandermonde’s identity is valid for n = N . Let us show that the identity is valid for n = N + 1.
*k
where the last line follows from the Vandermonde’s identity for n = N and Exercise 3.47(c).
b) Fix an arbitrary k ∈ {0, · · · , n + m}. Clearly, there are %n
j
&
subsets of Ω1 = {a1, · · · , an} consisting of j elements, and there are % m
k−j 3.49 First note that
(a1+ a2+ · · · + am)n=
*m j1,··· ,jn=1
aj1aj2. . . ajn.
For all n1, · · · , nm∈ {0, · · · , n} such that n1+ · · · + nm = n, define
Jn1,··· ,nm = {(j1, · · · , jn) : nkof the indices j1, · · · , jn are equal to k, for all k = 1, · · · , m}.
Note that the total number of elements in Jn1,··· ,nm is equal to % n
n1,··· ,nm
&
, since it’s the total number of ways to distribute n indices into m categories so that n1 of them are equal to 1, n2
of them are equal to 2, etc., nm of them are equal to m. Moreover, {(j1, · · · , jn) : jk∈ {1, · · · , m} for all k = 1, · · · , n} = +
n1,··· ,nm∈N ∪{0}:
n1+···+nm=n
Jn1,··· ,nm.
Therefore,
*m j1,··· ,jn=1
aj1aj2. . . ajn = *
n1,··· ,nm∈N ∪{0}:
n1+···+nm=n
' *
(j1,··· ,jn)∈Jn1,··· ,nm
aj1aj2. . . ajn
(
= *
n1,··· ,nm∈N ∪{0}:
n1+···+nm=n
' *
(j1,··· ,jn)∈Jn1,··· ,nm
an11an22. . . anmm (
= *
n1,··· ,nm∈N ∪{0}:
n1+···+nm=n
' n
n1, · · · , nm (
an11an22. . . anmm.
Therefore,
(a1+ a2+ · · · + am)n= *
n1,··· ,nm∈N ∪{0}:
n1+···+nm=n
' n
n1, · · · , nm
(
an11an22. . . anmm.
Advanced Exercises
3.50 For k = 0, · · · , n, the number of subsets of size k of a set, containing n elements, is equal to %n
k
&
. Thus, the total number of subsets of a set of n elements is equal to:
*n k=0
'n k (
=
*n k=0
'n k (
1k· 1n−k = (1 + 1)n= 2n, by the Binomial theorem.
3.51 To compute the number of distinct positive integer-valued solutions to x1+ · · · + xr = m, where r, m ∈ N , r ≤ m, imagine that there are m indistinguishable objects placed in a line and we want to subdivide them into r nonempty groups. Then to every positive integer-valued solution (x1, · · · , xr) of the equation x1+ · · · + xr = m, we can associate the distribution of objects of the form:
o · · · o
! "# $
x1
| o · · · o
! "# $
x2
| . . . | o · · · o
! "# $
xr
,
where “o” stands for the object in question, while “|” denotes a line separating adjacent groups.
Note that since there are r different groups, then there must be r − 1 separating lines, and those have to be placed in distinct spaces between the m objects. Note that any such distribution of objects is uniquely determined by the positions of separating lines and vice versa. Since there are m − 1 spots available (between the m objects) where the r − 1 separating lines can be placed (with no more than one line put between every pair of objects), then there are %m−1
r−1
&
ways to subdivide m indistinguishable objects into r groups of positive integer-valued sizes (x1, x1, · · · , xr). Moreover, there is a one-to-one correspondence between such subdivisions and the positive integer-valued solutions to the equation x1 + · · · + xr = m. Therefore, the total number of positive integer-valued solutions to x1+ · · · + xr= m is equal to %m−1
r−1
&
.
3.52 For r, m ∈ N , let n0(r, m) denote the number of non-negative integer-valued solutions to the equation x1+ · · · + xr = m and let n(r, m) be the number of positive integer-valued solutions
where the last equality holds by the Vandermonde’s identity (Exercise 3.48).
An even simpler solution is obtained by noting that the number of non-negative integer-valued solutions of x1 + · · · + xr = m is the same as the number of positive integer-valued solutions of y1+ · · · + yr= m + r (seen by letting yi = xi+ 1, i = 1, · · · , r). Thus, the answer is%m+r−1
r−1
&
, by Exercise 3.51.
3.53 a) %10+4−1
= 84, by Exercise 3.51.
3.54 %100+6−1
= 96, 560, 646. Refer to the solution to Exercise 2.81 and note that the number of outcomes in Ω is equal to the number of non-negative integer-valued solutions to the equation: x1+ x2+ x3+ x4+ x5+ x6 = 100. The required answer then follows by Exercise 3.52.
3.55 a) If the balls are all distinguishable and some urns are allowed to stay empty, then the number of possible outcomes is equal to Nn, since each of the n balls can be placed in any of the N (distinguishable) urns.
b) If the balls are indistinguishable and some urns are allowed to stay empty, then the number of possible outcomes coincides with the number of non-negative integer-valued solutions to the equation x1 + · · · + xN = n (where xk corresponds to the number of balls in urn #k, k = 1, · · · , N ), and is equal to %n+N −1
N−1
&
by Exercise 3.52.