There are $n$ identical red balls & $m$ identical green balls. The number of different linear arrangements consisting of "$n$ red balls but not necessarily all the green balls" is $\binom{x}{y}$. Find $x$ and $y$.
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Hint: Put one more red ball in front of the line. Now find the arrangements of the $(n+1)$ red balls and $m$ green balls. – Dec 06 '16 at 14:53
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The number is $$ \binom{n+m+1}{m}. $$ This is the number of linear arrangements of $n+1$ red balls and $m$ green balls. This number is equal to the number of arrangements you asked for given by the bijection "remove the rightmost red ball, as well as all green balls to the right of that one red ball", with inverse "add a red ball to the right of your arrangement, and then all unused green balls to the right of that".
Mees de Vries
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