Power set
2.5. MULTINOMIAL COEFFICIENTS 95 is 11. Hence the required number is the number of permutations
of the multiset M which is equal to the multinomial (4-nomial) coefficient
( 11 1,4,4,2
)
= 11!
1!4!4!2! = 34650.
For the second question, consider the four I’s as a single en- tity I’, that is, I′ ≡ IIII. Then we have the multiset M′ = {1.M,4.S,1.I′,2.P } of 8 elements. The number of permuta- tions of M′ is the 4-nomial coefficient ( 8
1,4,1,2
) which is equal to
8!
1!4!1!2! = 840. Therefore the number of permutations of the multi- setM in which the four I’s appear together is 840.
Hence, bythe subtraction rule, the required number is 34650− 840 = 33810.
Example 2.5.7 (Unordered partition): Find the number of unordered {|2,2, . . . ,{z 2}}
n2′s
partitions of the 2n set A = {1,2, . . . ,2n}. (For example, there are three {2,2,2} unordered partitions of the 4-set {1,2,3,4}, namely,
{ {1,2},{3,4} } ,{ {1,3},{2,4} } ,{ {1,4},{2,3} }).
The number of ( 2,2, . . . ,2 )
| {z }
n2′s
ordered partitions of A of the 2n- set is the multinomial coefficient ( 2n
2,2,...,2
) (by definition).
Eachunordered{|2,2, . . . ,{z 2}}
n2′s
partition{A1, A2, . . . , An}of the 2n-set A gives rise to n! ( 2,2, . . . ,2 )
| {z }
n2′s
ordered partitions of A (be- cause there are n! possible permutations of A1, A2, . . . , An.) Con- versely, the n! ( 2,2, . . . ,2 )
| {z }
n2′s
ordered partitions A1, A2, . . . , An of A define only oneunordered{|2,2, . . . ,{z 2}}
n2′s
partition{A1, A2, . . . , An} ofA (since we disregard the order of sets in the partition.) Hence
the number of unordered {|2,2, . . . ,{z 2}}
n2′s
partitions is ( 2n
2,2, . . . ,2 )
/n! = (2n)!
2nn!
Example 2.5.8 (Unordered partition):
Find the number of partitions of then-set A={1,2, . . . , n} into which j subsets can be partitioned such that there areki subsets havingi≥1 elements.
Solution: The number of subsets into which A should be par- titioned isj =k1+k2+· · · and the number of elements of the set A isn =k1+ 2k2+ 3k3+· · ·.
The number of ordered
k11′s
z }| { 1,1, . . . ,1,
k22′s
z }| { 2,2, . . . ,2,
k33′s
z }| { 3,3, . . . ,3, . . .
partitions is the multinomial coefficient
( n
1,1, . . . ,1,2,2, . . . ,2,3,3, . . . ,3,4, . . . )
= n!
1!k12!k23!k3· · ·. Eachunordered
k11′s
z }| { 1,1, . . . ,1,
k22′s
z }| { 2,2, . . . ,2,
k33′s
z }| { 3,3, . . . ,3, . . .
partition gives rise to k1!k2!k3!· · · ordered
k11′s
z }| { 1,1, . . . ,1,
k22′s
z }| { 2,2, . . . ,2,
k33′s
z }| { 3,3, . . . ,3, . . .
partitions (because there are k1! ways to permute subsets with one element, k2! ways to permute subsets with 2 elements etc., hence there are k1!k2!k3!· · · ways to permute all the subsets by
2.5. MULTINOMIAL COEFFICIENTS 97 the product rule (Proposition 2.2.2) and conversely. Hence the required number is
( n
1,1, . . . ,1,2,2, . . . ,2,3,3, . . . ,3,4, . . . )
/k1!k2!k3!· · ·= n!
1!k1k1!2!k2k2!3!k3k3!· · ·.
The following example illustrates a special case of the above example.
Example 2.5.9:
Find the number of unordered {1,1,2} partitions of the 4-set {1,2,3,4}.
By the formula of Example 2.5.8, the required number is
4!
(1!)22(2!)11! = 6. The six {1,1,2} unordered partitions Pi ={A1, A2, A3} for i= 1,2, . . . ,6 are given in the table:
Table 2.5: Six {1,1,2} partitions of the 4-set Partitions A1 A2 A3
P1 {1} {2} {3,4} P2 {1} {3} {2,4} P3 {1} {4} {2,3} P4 {2} {3} {1,4} P5 {2} {4} {1,3} P6 {3} {4} {1,2}
Example 2.5.10:
How many words of length 10 (that is 10-letter words) can be formed from 3 a’s, 3 b’s, 2 c’s and 2 d’s?
Note that 3+3+2+2=10. We shall use the bijective method.
Consider the 10-set A ={1,2, . . . ,10}.
We shall establish a one-to-one correspondence between the set of all ordered (3,3,2,2) partitions of the set A and the set B of all words of length 10 with exactly 3 a’s, 3 b’s, 2 c’s and 2 d’s.
Consider an ordered (3,3,2,2) partition (A1, A2, A3, A4) of the setA.We associate to the partition (A1, A2, A3, A4) a unique word wof length 10 belonging to the setB in the following way: Denote byw(i) the ith letter of the word w.
w(i) =
a if i∈A1
b if i∈A2 c if i∈A3 d otherwise
For example, to the partition (A1, A2, A3, A4) where A1 = {1,5,9, },A2 = {4,6,8},A3 = {2,7} and A4 = {3,10} we associate the wordacdbabcbad.
Conversely, given a wordwof length 10 belonging to the setB, we construct a unique ordered (3,3,2,2) partition (A1, A2, A3, A4) of the set A in the following way. Let i be an integer such that 1≤i≤10.
i∈
A1 if w(i) = a A2 if w(i) = b A3 if w(i) = c A4 otherwise
For example, to the word w = aabccdbabd the assigned partition isA1 ={1,2,8},A2 ={3,7,9},A3 ={4,5},A4 ={6,10}.
Therefore a one-to-one correspondence is found from the set of all ordered (3,3,2,2) partitions of the 10-set A onto the set B.
But the number of ordered (3,3,2,2) partitions of the 10-set A is the multinomial coefficient ( 10
3,3,2,2
). Hence (by the Proposition 2.2.1, the number of words of the set B is also equal to ( 10
3,3,2,2
), which is equal to 3!3!2!2!10! = 25200.
The above example motivates the following generalization and another interpretation of multinomial coefficients.
Theorem 2.5.2:
Consider an alphabetA ={a1, a2, . . . , ak}consisting ofk distinct
2.5. MULTINOMIAL COEFFICIENTS 99 letters. Letn1, n2, . . . nkbe knon-negative integers withn1+n2+
· · ·+nk =n. Then the number of words of length n that can be formed using the letter a1 exactly n1 times, using the letter a2 exactly n2 times, etc., and using the letter ak exactly nk times is the multinomial coefficient
( n
n1, n2, . . . , nk )
= n!
n1!n2!· · ·nk!.
Proof. It is enough if we imitate the proof of Example 2.5.10 by replacing the 10-set by the k-set A to establish a one-to-one cor- respondence between the ordered (n1, n2, . . . , nk) partitions of an n-set and the set of all words of length n from the alphabet A containing exactly n1 times the letter a1, containing exactly n2 times the letter a2, and finally containing exactly nk times the letter ak.
Example 2.5.11:
Find the number of words of length 4 (4-letter words) that can be formed from the letters a, b, c in which the letter a appears at most three times, b appears at most once and c appears at most twice.
First we enumerate all the possible ordered partitions of 4 into three parts in which the first part is ≤ 3 (corresponding to the maximum number of appearances of a), the second part is ≤ 1 (corresponding to the maximum number of appearances of b) and the third part is ≤ 2 (corresponding to the maximum number of appearances of c).
The possible partitions of the integer 4 with the above condi- tions are: (3,1,0),(3,0,1),(2,1,1),(2,0,2),(1,1,2).
Therefore, the number of words of length 4 with the imposed conditions is
( 4 3,1,0
) +
( 4 3,0,1
) +
( 4 2,1,1
) +
( 4 2,0,2
) +
( 4 1,1,2
)
which is equal to 4!
3!1!0!+ 4!
3!0!1!+ 4!
2!1!1!+ 4!
2!0!2!+ 4!
1!1!2! = 4 + 4 + 12 + 6 + 12 = 38.
Multinomial theorem:
The main use of the multinomial coefficient is the following gen- eralization of the binomial Theorem 2.4.1.
Theorem 2.5.3:
For any k real numbers a1, a2, . . . , ak and for any positive integer n the following relation is satisfied:
(a1+a2+· · ·+ak)n = ∑
n1,n2,...,nk≥0 n1+n2+···+nk=n
( n
n1, n2, . . . , nk )
an11an22· · ·ankk
Proof. First of all, observe that the expansion of (a1+a2+· · ·+ ak)n, is a homogeneous expression in a,a2, . . . , ak of degree n = n1+n2+· · ·+nk. Further, all possible terms of the forman11an22· · · ankk for n1, n2, . . . , nk ≥ 0 with n1+n2 +· · ·+nk = n are found in the expansion (see Observation 2.4.1). To get the expansion (a1+a2+· · ·+ak)n it is enough if we find the coefficient (that is the number of occurrences) of each terman11an22· · ·ankk.
Now consider the k-set of alphabet A = {a1, a2, . . . , ak}. A term of the forman11an22· · ·ankk in the expansion of (a1+a2+· · ·+ ak)n can be viewed as a word of length n1 +n2+· · ·+nk with the letter a1 appearing exactly n1 times, the letter a2 appearing exactly n2 times, etc., and finally the letter ak appearing exactly nk times. By Theorem 2.5.2, the number of such words is the multinomial coefficient
( n
n1, n2, . . . , nk )
Hence the relation is proved.
Example 2.5.12 (Multinomial theorem): Find the expansion of (a1+a2+a3)3.
The possible ordered partitions of the integer 3 into nonnega- tive integers are: (3,0,0),(0,3,0),(0,0,3),(2,1,0)(1,2,0),(2,0,1), (1,0,2),(0,2,1), (0,1,2),(1,1,1).
The partition (n1, n2, n3) corresponds to the terman11an22an33 in the expansion of (a1+a2+a3)3 and its coefficient is( 3
n1,n2,n3
). The
2.5. MULTINOMIAL COEFFICIENTS 101