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

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