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=(10!)/(2!(10-2)!)(8!)/(3!(8-3)!)(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}=(10!)/(2!3!5!)=2520`