Question

How would i solve a problem like this?

Answer 1

The number of combinations of size $k$ that you can make with $n$ items is given by the so-called binomial coefficient,

$\dbinom nk = \dfrac{n!}{k!(n-k)!}$

• $n!$ is the number of ways of permuting $n$ items.

• $(n-k)!$ is the number of ways of permuting all but $k$ of the $n$ items.

Dividing $n!$ by $(n-k)!$ then gives the number of ways of permuting only $k$ of the total $n$ items.

• $k!$ is the number of ways of permuting $k$ items.

Dividing $\frac{n!}{(n-k)!}$ by $k!$ then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

$\dbinom 63 = \dfrac{6!}{3!(6-3)!} = 20$

ways of selecting any 3 girls from the total 6 girls in the entire group of people.

There are

$\dbinom 72 = \dfrac{7!}{2!(7-2)!} = 21$

was of selecting any 2 boys from the total 7 boys.

Then there are

$\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}$

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

$\dbinom{13}5 = \dfrac{13!}{5!(13-5)!} = 1287$

Then the probability of selecting such a committee at random is

$\dfrac{\binom 63 \binom72}{\binom{13}5} = \dfrac{420}{1287} = \dfrac{140}{429} \approx 0.3263$