Answer to Question #148102 in Discrete Mathematics for Promise Omiponle

Question

How many ternary strings (i.e., the only allowable characters are 0, 1, and 2) of length 15 are there containing exactly four 0s, five 1s, and six 2s?

Answer 1

If M is a finite multiset, then a multiset permutation is an ordered arrangement of elements of M in which each element appears a number of times equal exactly to its multiplicity in M. An anagram of a word having some repeated letters is an example of a multiset permutation. If the multiplicities of the elements of M (taken in some order) are ${\displaystyle m_{1}} , {\displaystyle m_{2}} , ..., {\displaystyle m_{l}}$ and their sum (that is, the size of M) is n, then the number of multiset permutations of M is given by the multinomial coefficient,

${\displaystyle {n \choose m_{1},m_{2},\ldots ,m_{l}}={\frac {n!}{m_{1}!\,m_{2}!\,\cdots \,m_{l}!}}={\frac {\left(\sum _{i=1}^{l}{m_{i}}\right)!}{\prod _{i=1}^{l}{m_{i}!}}}.}$

In our case, the number of ternary strings of length 15 are there containing exactly four 0s, five 1s, and six 2s is

$\frac {15!}{4!\,5!\,6!}=\frac {15\cdot 14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7}{(2\cdot3\cdot4)\,(2\cdot3\cdot4\cdot5)}=630,630$