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
 A: My answer does indeed match yours.  Here below is my approach (which is much less tedious for arithmetic).
To have at least one voided suit we approach via inclusion-exclusion in the following way:
$|\text{at least one void}| = |\text{one guaranteed void}|-|\text{two guaranteed void}| + |\text{three guaranteed void}| - |\text{four guaranteed void}|$
To find, say $|\text{two guaranteed void}|$, pick which two suits are guaranteed to be voided ($\binom{4}{2}$ number of ways of choosing) and then pick which cards appear such that none are from the guaranteed voided suits ($\binom{52}{26}$ number of ways).  Note that we do not bother finding the amount for exactly two voided suits as the overcounting is taken care of through the inclusion-exclusion process and the answer to the question of exactly two voided suits is irrelevant to our original question.
$|V_{\geq 1}| = \binom{4}{1}\binom{39}{13}-\binom{4}{2}\binom{26}{13}+\binom{4}{3}\binom{13}{13}-\binom{4}{4}\binom{0}{13}$
The last term of course is zero since it is impossible to have four voided suits in a hand of thirteen cards.
my answer=your answer=$32~427~298~180$
The negation of "at least one voided suit" is "strictly fewer than one voided suit" which in this case means no voided suits.  This can be found by the fact that these two form a partition of our sample space of all ways of picking thirteen cards.
$|V_0|=\binom{52}{13}-\binom{4}{1}\binom{39}{13}+\binom{4}{2}\binom{26}{13}-\binom{4}{3}\binom{13}{13} = 602~586~261~420$
A: 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.
