Combinatorics, how to pick X of one item, and Y of another out of Z total items?

Question

Let's say I have several kinds of bricks. Red bricks, yellow bricks, and blue bricks.

If I have infinite bricks, but am only selecting a group of 15 bricks, what is the chance I pick 7 red, 5 yellow, and 3 blue bricks in those 15?

I'm thinking there's $3^{15}$ different sequences of choosing those bricks, and 15 choose 7 ways of selecting the red bricks, 8 choose 5 ways of selecting the yellow bricks, and 3 choose 3 ways of selecting the yellow which I would multiply?

So $$\frac{{15\choose 7} {8\choose 5} {3\choose3}}{3^{15}}$$

But I was thinking I could also go about choosing 3 out of 15 first, rather than 7 out of 15, which would change the number.

Answer 1

Yes, that is correct reasoning, and no, changing the order of the brick colors would not change the value; in all cases there are $$\frac{15!}{3!\cdot5!\cdot7!}=360360$$ possible ways to get that pile of bricks. A somewhat better tool here is the Multinomial Coefficient, which more directly handles more than two colors of brick.

EDIT: I guess more detail would be helpful.

When you did this with binomial coefficients, you did $\binom{15}{7}\binom{8}{5}\binom{3}{3}$. Which, if you do it out, gives

$$\frac{15!}{7!\cdot8!}\cdot\frac{8!}{5!\cdot3!}\cdot\frac{3!}{3!\cdot0!}$$

Which cancels...

$$\require{cancel}\frac{15!}{7!\cdot\cancel{8!}}\cdot\frac{\cancel{8!}}{5!\cdot\cancel{3!}}\cdot\frac{\cancel{3!}}{3!}=\frac{15!}{7!\cdot5!\cdot3!}$$

This cancellation happens no matter which order you apply your three (or more!) counts.

Answer 2

You are right on track with your reasoning!

Except the $3^{15}$ on the bottom doesn't belong. If you were to leave that on the bottom you would be left with $$\frac{360360}{14348907}=0.00812556665$$ Which would be way to small of an answer.