Combinatorics
2.6 Generating Functions
2.6.2 Counting with Repetition
Then, shalt thou count to three, no more, no less. Three shalt be the number thou shalt count, and the number of the counting shall be three. Four shalt thou not count, nor either count thou two, excepting that thou then proceed to three. Five is right out.
— Monty Python and the Holy Grail Suppose there is an inexhaustible supply of each ofndifferent objects. How many ways are there to selectmobjects from thendifferent objects, if you are allowed to select each object as many times as you like?
Letan,mdenote this number. Evidently, for fixedn, the generating function for {an,m}m≥0is
Gn(x) =
1 +x+x2+· · ·n
= 1
1−x n
,
since the sum is just a geometric series inx. This raises questions on convergence, for this formula is valid only for|x|<1. We largely ignore these analytic issues, since we treat generating functions as formal series.
Thus, to find a formula foran,m, we must find the coefficient ofxminGn(x).
Let us consider a more general problem. Letf(x) = (1 +x)α, whereαis a real number. Thenf(0) =α,f(0) =α(α−1), and in general,f(k)(0) =αk. Therefore, the Maclaurin series forf(x)is
(1 +x)α=
k≥0
αk k!xk.
Define the generalized binomial coefficient by α
k
=
αk/k! ifk≥0,
0 ifk <0. (2.33) Note thatα
k
equals the ordinary binomial coefficient wheneverαis a nonnega- tive integer. We have the following theorem.
Theorem 2.7 (Generalized Binomial Theorem). If|x|<1orαis a nonnegative integer, then
(1 +x)α=
k
α k
xk. (2.34)
The proof of convergence may be found in many analysis texts, where it is often proved as a consequence of Bernstein’s theorem on convergence of Taylor series (see for instance [11]). We do not supply the proof here.
Before solving our problem concerning selection with unlimited repetition, we note a useful identity for generalized binomial coefficients.
Negating the Upper Index. Ifαis a real number andkis an integer, then α
k
= (−1)k
k−α−1 k
. (2.35)
Proof. Fork <0, the identity is clear. Fork≥0, we have α
k
= 1 k!
k−1 i=0
(α−i).
Reindex this product, replacing eachibyk−1−i, to obtain α
k
= 1 k!
k−1 i=0
(α−(k−i−1))
= (−1)k k!
k−1 i=0
(k−1−i−α)
= (−1)k
k−α−1 k
.
We may now solve our problem of determiningan,m. We compute Gn(x) = (1−x)−n
=
m
−n m
(−x)m
=
m
n+m−1 m
xm,
and therefore the number of ways to selectmobjects from a collection ofndif- ferent objects, with repetition allowed, is
an,m=
n+m−1 m
. (2.36)
For example, the number of five-card poker hands that can be dealt from a stack of five or more decks is56
5
= 3 819 816.
Finally, suppose we lay all52cards of a standard deck face up on a table. How many ways can we place five identical poker chips on the cards if we allow more than one chip to be placed on each card? To solve this, notice that each possible placement of chips corresponds to a hand of five cards, where repeated cards are allowed: Ifkchips lie on a particular card, place that card into the handktimes.
Further, every such five-card hand can be represented by a judicious placement of chips. Therefore, the answer is the same as that of the previous example,56
5
. In general, the number of ways to placemidentical objects intondistinguish- able bins is the same as the number of ways to selectmobjects from a set ofn objects with repetition allowed: The answer to both problems isn+m−1
m
.
Exercises
1. Prove the addition identity for generalized binomial coefficients: Ifαis a real number andkis an integer, then
α k
= α−1
k
+ α−1
k−1
.
2. Prove the absorption/extraction identity for generalized binomial coeffi- cients: Ifαis a real number andkis a nonzero integer, then
α k
=α k
α−1 k−1
.
3. Prove the cancellation identity for generalized binomial coefficients: Ifαis a real number andkandmare integers, then
α k
k m
= α
m
α−m k−m
.
4. Prove the parallel summation identity for generalized binomial coefficients:
Ifαis a real number andnis an integer, then n
k=0
α+k k
=
α+n+ 1 n
.
5. Suppose that an unlimited number of jelly beans is available in each of five different colors: red, green, yellow, white, and black.
(a) How many ways are there to select twenty jelly beans?
(b) How many ways are there to select twenty jelly beans if we must select at least two jelly beans of each color?
6. A catering company brings fifty identical hamburgers to a party with twenty guests.
(a) How many ways can the hamburgers be divided among the guests, if none is left over?
(b) How many ways can the hamburgers be divided among the guests, if every guest receives at least one hamburger, and none is left over?
(c) Repeat these problems if there may be burgers left over.
7. A zodiac sign is one of twelve constellations that the sun travels through (from the vantage point of the earth) over the course of a year. Each person has a zodiac sign based on the position of sun on their birth date. The astro- logical configuration of a party withnguests is a list of twelve numbers that records the number of guests with each sign, so the first number records the number of people with the sign Capricorn, the second, Aquarius, . . . , the last, Sagittarius.
(a) How many different astrological configurations are possible forn= 100?
(b) How many astrological configurations are possible for n = 100, if each component is at least5?
8. Two lottery systems are proposed for a new state lottery. In the first sys- tem, players select six different numbers from{1,2, . . . ,50}. In the second system, players select six numbers from{1,2, . . . ,45}, and may select any number as many times as they want. (In the second system, each ball se- lected in the lottery drawing is replaced before another ball is selected.) Which system has more possible tickets?
9. Suppose100identical tickets for rides are distributed among40children at a carnival.
(a) How many ways can the tickets be distributed, if each child receives at least two tickets, and all the tickets are distributed?
(b) How many ways can the tickets be distributed, if each child receives at least one ticket, and some tickets may be left over?
(c) Suppose one child has twelve tickets, and each ticket may be used on any of six different rides. How many ways can the child spend her tickets, if she can choose any ride any number of times, and the order of choice is unimportant?