The number of ways of choosing `2` prefects from `5` is `C_2^5 =frac{5!}{2! times 3!} = 10`

The number of ways of choosing `6` non-prefects from `10` is

`C_6^10 =frac{10!}{6! times 4!} = 210`

(a) Number of possible committees with exactly `2` prefects:

`C_2^5 times C_6^10 = 10 times 210 = 2100`

(b) Number of committees with `3` prefects:

`C_3^5 times C_5^10 ` `= frac{5!}{3! times 2!} times frac{10!}{5! times 5!}` `= 2520`

Number of committees with `4` prefects:

`C_4^5 times C_4^10` `= frac{5!}{4! times 1!} times frac{10!}{4! times 6!}` `= 1050`

Number of committees with `5` prefects:

`C_5^5 times C_3^10` `= frac{5!}{5! times 0!} times frac{10!}{3! times 7!}` `= 120`

So the number of committees with at least `2` prefects is:

`2100 + 2520 + 1050 + 120 = 5790`

The problem with the method used above is that if we have many (say `20`) to count, it would become very tedious. So we look at another way of doing it.

If we find the number of committees with `0` prefects and `1` prefect, and subtract this from the total number of committees, we will have the number with at least `2`:

Number of committees with `0` prefects:

`C_8^10 =frac{10!}{8! times 2!} = 45`

Number of committees with `1` prefect:

`C_1^5 times C_7^10 = frac{5!}{1! times 4!} times frac{10!}{7! times 3!} = 600`

The total number of committees is:

`C_8^15 =frac{15!}{8!(15 - 8)!} = frac{15!}{8! times 7!} = 6435`

So the number with at least `2` prefects is given by:

`6435 − 45 − 600 = 5790`