Suppose we have a deck of $32$ cards with $8$ cards of each of the four suits. In how ways we can choose six cards such that there are cards of exactly three different suits among the chosen cards?
I believe inclusion-exclusion principle is the way to solve it, where we first count the number of total ways to choose $6$ cards out of $32$ (which is $\binom{32}{6}$), then exclude the number of combinations where exactly two of the suits are missing (which is $\binom{4}{2}\binom{16}{6}$) and then by inclusion-exclusion formula add the combinations where all three suits are missing (which is $\binom{4}{3}\binom{8}{6}$). The number of combinations of all $4$ suits missing is, of course, zero.
My question is - where is my logic incorrect? I know it is, but can't seem to spot the error.
$\binom{32}{6}$
. If you want to display the formula on its own line, type$$\binom{32}{6}$$
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