Suppose there are $300$ colored balls in an urn, with $100$ each of the colors red, blue, and green. How many ways are there to select $14$ balls from the urn if there must be at least $3$ but at most $10$ blue balls? Assume the order of selection does not matter.
If $b$ and $r$ are the number of blue and red balls taken from the urn, respectively, then I believe the answer to be
$$\sum_{b=3}^{10}\binom{100}b\sum_{r=0}^{14-b}\binom{100}r\binom{100}{11-r}$$
which apparently has a value of $2230480721192909724511239806000$. I obtained the above sum by generalizing cases in which $b$, then $r$, are fixed. (E.g. if $b=3$, that leaves us with $11$ remaining balls, all of which could be red and none of which would be green, etc.)
Is this correct? Next, how would I go about verifying this using a generating function?
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