There are 20 red, 20 green and 20 blue balls in an urn.
In how many different ways can 10 balls be selected?
How many ways are there if there are 6 red balls instead of 20?
I would think to solve this with the binomial coefficient (i.e. 60 choose 10) but this seems far too simplistic for a question on an otherwise very difficult assignment. Is there something I'm missing?
If you use stars and bars, the questions is how man ways can $10$ indistinguishable balls be place in $3$ buckets labelled "red," "green," and "blue" and the answer is $$\binom{10+3-1}{3-1} = \binom{12}{2}$$
For the second part of the question, we have the restriction that at most $6$ balls can be placed in the bucked marked "red," so we need to subtract the number of solutions to the first question with at least "7" balls in the red bucket. The way to do this is by placing $7$ balls in the red bucket to begin with, and then distribute the remaining $3$ balls in the $3$ buckets.
Can you finish from here?