Combinatorics
2.3 Multinomial Coefficients
10. In the Virginia lottery game Win For Life, an entry consists of a selection of six different numbers between1and42, and each drawing selects seven different numbers in this range. How many different entries can match at least three of the drawn numbers?
11. The state of Florida administers several lottery games. In Florida Lotto, a player picks a set of six numbers between1and53. In Fantasy5, a gambler chooses a set of five numbers between1and36. In which game is a player more likely to match at least two numbers against the ones drawn?
multinomial coefficient n
k1,k2,...,km
to be the number of ways to placek1of the objects in box1,k2in box2, and so on, without regard to the order of the objects in each box. Then an argument similar to our analysis for trinomial coefficients shows that
n k1, . . . , km
= n
k1
n−k1
k2
n−k1−k2
k3
· · ·
·
n−k1− · · · −km−2
km−1
= n!
k1!k2!· · ·km!.
(2.17)
Multinomial coefficients often arise in a natural way in combinatorial problems.
While we can always reduce questions about multinomial coefficients to prob- lems about binomial coefficients or factorials by using (2.17), it is often useful to handle them directly. We derive some important formulas for multinomial coef- ficients in this section. These generalize some of the statements about binomial coefficients from Section 2.2. We begin with a more general formula for expand- ing multinomial coefficients in terms of factorials.
Expansion. Ifnis a nonnegative integer, andk1, . . . ,kmare integers satisfying k1+· · ·+km=n, then
n k1, . . . , km
=
⎧⎨
⎩ n!
k1!· · ·km! if eachki≥0,
0 otherwise.
(2.18)
Taking n
k1,...,km
= 0when at least one of thekiis negative is certainly sensible, since it is impossible to place a negative number of objects in a box.
Second, it is clear that rearranging the numbersk1, . . . ,km does not affect the value of the multinomial coefficient n
k1,...,km
, since this just corresponds to relabeling the boxes. We can state this in the following way.
Symmetry. Supposeπ(1), . . . ,π(m)is a permutation of{1, . . . , m}. Then n
k1, . . . , km
=
n kπ(1), . . . , kπ(m)
. (2.19)
Third, we can observe a simple addition law. Letαbe one of the objects from the set ofn. It must be placed in one of the boxes. If we placeαin box1, then there are n−1
k1−1,k2,...,km
ways to arrange the remainingn−1objects to create a valid arrangement. If we setαin box2, then there are n−1
k1,k2−1,k3,...,km
to complete the assignment of objects to boxes. Continuing in this way, we obtain the following identity.
Addition. Ifnis a positive integer andk1+· · ·+km=n, then n
k1, . . . , km
=
n−1 k1−1, k2, . . . , km
+
n−1 k1, k2−1, k3, . . . , km
+· · ·+
n−1
k1, k2, . . . , km−1, km−1
.
(2.20)
In the last section, the addition identity form = 2produced Pascal’s triangle for the binomial coefficients. We can use a similar strategy to generate a geomet- ric arrangement of the trinomial coefficients whenm = 3, which we might call Pascal’s pyramid. The top level of the pyramid corresponds ton = 0, just as in Pascal’s triangle, and here we place a single1, for 0
0,0,0
. The next level holds the numbers forn= 1, and we place the three1s in a triangular formation, just below then= 0datum at the apex, for the numbers 1
1,0,0
, 1
0,1,0
, and 1
0,0,1
. In general, we use the addition formula (2.20) to compute the numbers in leveln from those in leveln−1, and we place the value of n
a,b,c
in levelnjust below the triangular arrangement of numbers n−1
a−1,b,c
, n−1
a,b−1,c
, and n−1
a,b,c−1
in level n−1. Figure 2.1 shows the first few levels of the pyramid of trinomial coefficients.
Here, the position of each number in levelnis shown relative to the positions of the numbers in leveln−1, each of which is marked with a triangle ().
1
1
1 1
1
2 2
1 2 1
n= 0 n= 1 n= 2
1
3 3
3 6 3
1 3 3 1
1
4 4
6 12 6
4 12 12 4
1 4 6 4 1
n= 3 n= 4
FIGURE 2.1. The first five levels of Pascal’s pyramid.
We can use the addition identity to obtain an important generalization of the binomial theorem for multinomial coefficients.
The Multinomial Theorem. Ifnis a nonnegative integer, then (x1+· · ·+xm)n=
k1+···+km=n
n k1, . . . , km
xk11· · ·xkmm. (2.21)
Here, the notation
k1+···+km=nmeans that the sum extends over all integer m-tuples(k1, . . . , km)whose sum isn. Of course, there are infinitely many such m-tuples, but only finitely many produce a nonzero term by the Expansion iden- tity, so this is in effect a finite sum. We prove (2.21) for the casem = 3; the general case is left as an exercise.
Proof. The formula
(x+y+z)n=
a+b+c=n
n a, b, c
xaybzc (2.22) certainly holds forn= 0, so suppose that it is valid forn. We compute
(x+y+z)n+1= (x+y+z)
a+b+c=n
n a, b, c
xaybzc
=
a+b+c=n
n a, b, c
xa+1ybzc+
a+b+c=n
n a, b, c
xayb+1zc
+
a+b+c=n
n a, b, c
xaybzc+1
=
a+b+c=n+1
n a−1, b, c
+
n a, b−1, c
+
n a, b, c−1
xaybzc
=
a+b+c=n+1
n+ 1 a, b, c
xaybzc,
so (2.22) holds for alln≥0.
Some additional formulas for multinomial coefficients are developed in the ex- ercises. Some of these may be obtained by selecting particular values forx1, . . . , xmin the multinomial theorem.
We close this section by describing a common way that multinomial coeffi- cients appear in combinatorial problems. Suppose we need to count the number of ways to order a collection ofnobjects. If all the objects are different, then the answer is simplyn!, but what if our collection includes some duplicate objects?
Such a collection is called a multiset. Certainly we expect fewer different arrange- ments when there are some duplicate objects. For example, there are just six ways
to line up four poker chips, two of which are red and the other two blue:rrbb, rbrb,rbbr,brrb,brbr, andbbrr.
Suppose we have a multiset of sizenthat includes exactlyk1identical copies of one object,k2 instances of another, and so on, ending withkmduplicates of the last object, sok1+· · ·+km=n. In any ordering of thesenobjects, we may rearrange thekicopies of objectiin any way without disturbing the arrangement.
Since we can do this for any of themobjects independently, it follows that each distinct ordering of the items occursk1!k2!· · ·km!times among then!ways that one could arrange the objects if they had been distinguishable. Therefore, the number of distinct arrangements of our multiset is
n!
k1!· · ·km! =
n k1, . . . , km
.
We could also obtain this formula by using our first combinatorial model for the multinomial coefficients. Suppose we haven ping-pong balls, numbered1 throughn, andmboxes, each labeled with a different object from our multiset.
The number of ways to distribute the balls among the boxes, withk1in box1,k2in box2, and so on, is the multinomial coefficient n
k1,...,km
. But each arrangement corresponds to an ordering of the elements of our multiset: The numbers in boxi indicate the positions of objectiin the listing.
We have thus answered the analogue of Problem 1 from Section 2.1 for multi- sets. We can also study a generalization of Problem 2: How many ways are there to make an ordered list ofrobjects from a multiset ofnobjects, if the multiset compriseski copies of objectifor1 ≤ i ≤ m? Our approach to this problem depends on theki andr, so we’ll study an example. Suppose a contemplative resident of Alma, Alabama, wants to know the number of ways to rearrange the letters of her home town and state, ignoring differences in case. There are eleven letters in all: six As, one B, two Ls, and two Ms, so she computes the total number to be 11
6,1,2,2
=6!2!2!11! = 13 860.
Suppose she also wants to know the number of four-letter sequences of letters that can be formed from the same string,ALMAALABAMA, like the ones in the list that open this section, only they do not have to be English words. This is the multiset version of Problem 2 withn= 11,r= 4,m = 4,k1= 6,k2 = 1, andk3=k4 = 2. We can solve this by constructing each sequence in two steps:
first, select four elements from the multiset; second, count the number of ways to order that subcollection. We can group the possible sub-multisets according to their pattern of repeated elements. For example, consider the subcollections that have two copies of one object, and two copies of another. Denote this pattern by wwxx. There are3
2
= 3ways to select values forwandx, since we must pick two of the three lettersA,L, andM. Each of these subcollections can be ordered in any of4
2,2
= 6ways, so the pattern wwxx produces3·6 = 18 possible four-letter sequences in all. There are five possible patterns for a four-element multiset, which we can denotewwww,wwwx,wwxx,wwxy, andwxyz. The analysis of each one is summarized in the following table.
Pattern Sub-multisets Orderings per sub-multiset Total
wwww 1 1 1
wwwx 3
1
4
3,1
12
wwxx 3
2
4
2,2
18
wwxy 3
1
3
2
4
2,1,1
108
wxyz 1 4! 24
Summing the values in the rightmost column, we find that there are exactly163 ways to form a four-letter sequence from the letters in Alma, Alabama.
Exercises
1. Prove the addition identity for multinomial coefficients (2.20) by using the expansion identity (2.18).
2. For nonnegative integersa,b, and c, letP(a, b, c)denote the number of paths in three-dimensional space that begin at the origin, end at(a, b, c), and consist entirely of steps of unit length, each of which is parallel to a coordinate axis. Prove thatP(a, b, c) =a+b+c
a,b,c
.
3. Prove the multinomial theorem (2.21) for an arbitrary positive integerm.
4. Prove the following identities for sums of multinomial coefficients, if m andnare positive integers.
(a)
k1+···+km
n k1, . . . , km
=mn.
(b)
k1+···+km
n k1, . . . , km
(−1)k2+k4+···+k2 =
0 ifm= 2, 1 ifm= 2+ 1.
5. Prove that ifnis a nonnegative integer andkis an integer, then
j
n j, k, n−j−k
= 2n−k n
k
.
6. Prove the multinomial theorem for falling factorial powers, (x1+· · ·+xm)n=
k1+···+km=n
n k1, . . . , km
x1k1· · ·xmkm,
and for rising factorial powers, (x1+· · ·+xm)n=
k1+···+km=n
n k1, . . . , km
xk11· · ·xkmm. You may find it helpful to consider the trinomial case first.
7. Use a combinatorial argument to establish the following analogue of Van- dermonde’s convolution for trinomial coefficients. Ifmandnare nonneg- ative integers, anda+b+c=m+n, then
α+β+γ=m
m α, β, γ
n
a−α, b−β, c−γ
=
m+n a, b, c
.
8. State an analogue of Vandermonde’s convolution for multinomial coeffi- cients, and use a combinatorial argument to establish it.
9. Compute the number ofr-letter sequences that can be formed by using the letters in each location below, for each given value ofr. Ignore differences in case.
(a) Bug Tussle, TX:r= 3,r= 4,r= 11.
(b) Cooleemee, NC:r= 4,r= 10,r= 11.
(c) Oconomowoc, WI:r= 4,r= 11,r= 12.
(d) Unalaska, Alaska:r= 3,r= 4,r= 14.
(e) Walla Walla, WA:r= 4,r= 5,r= 12.
10. Certainly there are more four-letter sequences that can be formed by using the letters in Bobo, Mississippi, than can be formed by using the letters in Soso, Mississippi. Is the difference more or less than the distance between these two cities in miles, which is267?
11. A band of combinatorial ichthyologists asserts that the number of five-letter sequences that can be formed using the letters of the Hawaiian long-nosed butterfly fish, the lauwiliwilinukunuku’oi’oi, is more than twice as large as the number of five-letter sequences that can be created using the name of the state fish of Hawaii, the painted triggerfish humuhumunukunukuapua’a.
Prove or disprove their claim by computing the exact number in each case.