<s>[INST] Card Draw question What are the ways to draw 13 cards from a pack of 52 cards such that 
(a) the hand is void in at least one suit, (b) the hand is not void in any suit.(“void in a suit” means having no cards of that suit)
Current approach
I am thinking on following lines:
Case-1: All cards from exactly 1 suit $= {4 \choose 1} * {13 \choose 13} = 4$ ways
Case-2: Cards from exactly 2 suits = ${4 \choose 2} * [{26 \choose 13}  - {2 \choose 1} * {13 \choose 13}] = 62,403,588$ ways
Case-3: Cards from exactly 3 suits = ${4 \choose 3} * [{39 \choose 13} - {3 \choose 2} * [{26 \choose 13} - {2 \choose 1} * {13 \choose 13}] - {3 \choose 1} * {13 \choose 13}] = 32,364,894,588$ ways
Case-4: Cards from exactly 4 suits = ${4 \choose 4} * [{52 \choose 13} - {4 \choose 3} * [{39 \choose 13} - {3 \choose 2} * [{26 \choose 13} - {2 \choose 1} * {13 \choose 13}]] - {4 \choose 2} * [{26 \choose 13} - {2 \choose 1} * {13 \choose 13}] - {4 \choose 1} * {13 \choose 13}] = 602,586,261,420$ ways
I have no ways to verify answer for Case-4. However as expected, answer for (a) matches sum case-1 + case-2 + case-3. I am sure there must be a simpler and more direct way to arrive at these numbers. Any insights. Many Thanks
 [/INST] For (a), select the 3 suits represented out of 4, and then select 13 cards out of the remaining $52 - 13 = 39$, for a total of $\binom{4}{3} \cdot \binom{39}{13} = 32489701776$. For (b), it is all posibilities (taking 13 card out of 52) less the ones missing a suit. Part (a) uses at most 3 suits, by a similar reasoning:


*

*At most one suit: $\binom{4}{1} \binom{13}{13}$

*At most two suits: $\binom{4}{2} \binom{2 \cdot 13}{13}$

*At most three suits: $\binom{4}{3} \binom{3 \cdot 13}{13}$


Then exactly 2 suits is (at most 3) - (at most 2), i.e.,
$$
\binom{4}{1} \binom{13}{13} - \binom{4}{2} \binom{2 \cdot 13}{13}  = 32427298176
$$
This is essentially inclusion-exclusion.
Thanks to JMoravitz for catching a dumb mistake. I plead not enough coffee.
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