TITLE: Verifying a combinatorics problem: choosing 54 numbers out of 108?
QUESTION [5 upvotes]: I have the problem and a proposed solution, however I want to verify that it's correct, as sometimes I struggle visualizing combinatorics.
Let's say you have a grid of 108 squares. You have 54 blue and 54 red balls. How many different ways can you place the blue and red balls? 
My proposed solution: assign each square a number 1-108. That leaves us 108 numbers. Because it's the same number of red and blue balls, you can just ask the question: "how many different ways can I place blue or red balls on this grid?" or the same as saying $108 \choose 54$ which is over $2 * 10^{31}$.
Is this correct? Out of curiosity, how would you approach it if there weren't the same number of red and blue balls?

REPLY [6 votes]: In general, if you have $n$ positions, with $k$ blue balls and $n-k$ red balls, then the number of ways is $\binom{n}{k} = \binom{n}{n-k}$. To see this, it suffices to choose how to place the $k$ blue balls, which is $\binom{n}{k}$; the red balls will just occupy the remaining places. A symmetric justification is that there are $\binom{n}{n-k}$ ways to place the red balls; the blue balls will then occupy the remaining places. Noting that these two approaches give the same answer, that is, $\binom{n}{k} = \binom{n}{n-k}$, is a matter of writing out the definition of $\binom{n}{k} = \frac{n!}{k!(n-k)!} = \binom{n}{n-k}$.