Question

How many 10-digit ternary strings are there that contain exactly two 0s, three 1s, and five 2s?

Answer 1

There are $\dbinom{10}2$ ways of picking 2 of the 10 available positions for a 0. 8 positions remain.

There are $\dbinom83$ ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's $\dbinom55=1$ way of doing that.

So we have

$\dbinom{10}2\dbinom83\dbinom55=\dfrac{10!}{2!(10-2)!}\dfrac{8!}{3!(8-3)!}\dfrac{5!}{5!(5-5)!}=2520$

The last expression has a more compact form in terms of the so-called multinomial coefficient,

$\dbinom{10}{2,3,5}=\dfrac{10!}{2!3!5!}=2520$