3 types of balls in an urn but only a few ways

Question

There are $15$ Black, $15$ White and $15$ Brown balls in an urn. In how many different ways can $5$ balls be selected? How many ways are there if there are $3$ black balls instead of $15$?

My approach

Here for part one I thought of doing $(15+15+15)\choose 5$ which is $45\choose5$. However then I thought of a different idea, which is that when we see it this way, there are only a few ways that can be formed:

1: $5-0-0 $

2: $5-1-0 $

3: $5-0-1 $

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.

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n: $0-0-5$

so would it be $n$ number of ways or $45\choose 5$ . And if it is $n$, then how would I find n?

Answer 1

You are expected to consider all the balls of a color identical. As long as there are at least five of each it doesn't matter how many there are because you won't run out of any color. It is not clear if order matters. If it does, you have three choices at each draw and $3^5$ overall. If not, you have a stars and bars problem to find three numbers allowing $0$ that add to $5$. This is ${5+3-1\choose 3-1}=21$