# A bag contains 4 red, 6 white, and 7 blue marbles. How many ways can 3 red, 2 white, and 4 blue marbles be chosen?

Oct 17, 2017

$1260$

#### Explanation:

Assuming that the marbles are all identical, we're looking for the number of different orders that you could pick the marbles in. Since order matters in this case, we're looking for permutations, but since some of the items we're permuting are the same, we have to modify the formula a bit.

If we were looking at how many ways that 9 different marbles could be chosen, then the answer would simply be 9! (factorial), since there are 9 possibilities for the first marble, times 8 for the second, and so on.

However, some of these marbles are identical. The formula for a permutation like this is:

(n!)/(n_1! xx n_2! xx n_3! xx cdots)

Where ${n}_{1}$ is the number of identical objects of one type, ${n}_{2}$ is the number of identical objects of another type, and so on.

That might seem confusing, so let's go ahead and just use this formula to solve the problem. It'll become much clearer once we start plugging numbers in.

We have $\textcolor{\lim e g r e e n}{9}$ marbles to permute. $\textcolor{red}{3}$ of them are red, $\textcolor{\mathmr{and} a n \ge}{2}$ of them are white, and $\textcolor{b l u e}{4}$ of them are blue. Therefore, the number of ways to choose the marbles is:

(color(limegreen)9!)/(color(red)3! xx color(orange)2! xx color(blue)4!)

This is our answer! Now, all we have to do is simplify it.

$\frac{\textcolor{\lim e g r e e n}{9} \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{\textcolor{red}{3} \times 2 \times 1 \times \textcolor{\mathmr{and} a n \ge}{2} \times 1 \times \textcolor{b l u e}{4} \times 3 \times 2 \times 1}$

$\frac{\textcolor{\lim e g r e e n}{9} \times 8 \times 7 \times 6 \times 5 \times \cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\textcolor{red}{3} \times 2 \times 1 \times \textcolor{\mathmr{and} a n \ge}{2} \times 1 \times \cancel{\textcolor{b l u e}{4}} \times \cancel{3} \times \cancel{2} \times \cancel{1}}$

$\frac{\textcolor{\lim e g r e e n}{9} \times 8 \times 7 \times \cancel{6} \times 5}{\cancel{\textcolor{red}{3}} \times \cancel{2} \times 1 \times \textcolor{\mathmr{and} a n \ge}{2} \times 1}$

$\frac{\textcolor{\lim e g r e e n}{9} \times \left(4\right) \cancel{8} \times 7 \times 5}{1 \times \cancel{\textcolor{\mathmr{and} a n \ge}{2}} \times 1}$

$9 \times 4 \times 7 \times 5$

$1260$