How many 5-cards poker hands are there containing at least 3 of the 4 suits?
I don't how know how to properly handle this question.
There 13 ranks of each card, and 4 suits
My Attempt:
Begin with the first 3: C(13,3) X 4 ways to choose suits
Then pick last 3 cards of anything:
C(49,2)
My solution: 1,345,344
Thanks,
Best Answer
EDIT ( thanks to Thomas ) ****
There are $\binom{52}{5}$ possible 5 card hands $$N_{tot} = \binom{52}{5}$$
for the number of hands containing just one suit there are 4 ways to choose the suit and $\binom{13}{5}$ ways to choose 5 cords from that suit
$$N_{1} = 4\binom{13}{5}$$
For the number of hands containing exactly 2 suits there are $\binom 42$ ways to choose the 2 suits
for each pair of suits there are $\binom{26}{5}$ 5 card hands, but that includes $\binom {13}{5}$ one suit hands from each of the 2 suits. $$N_{2} = \binom 42 \left(\binom{26}{5}-2\binom{13}{5} \right)$$ so $$N_{\text{at least 3}}=N_{tot}-N_1-N_2 = \binom {52}{5} - 4\binom{13}{5} - \binom 42 \left(\binom{26}{5}-2\binom{13}{5} \right) $$
* Original (incorrect) Answer *
P(all from one suit) $$ = \frac {4\binom{13}{5}}{\binom{52}{5}} $$
P(all from two suits) $$ = \frac {\binom 42 \binom{26}{5}}{\binom{52}{5}} $$
P(all from at least 3 suits) $$ = 1-\frac {4\binom{13}{5}}{\binom{52}{5}} - \frac {\binom 42 \binom{26}{5}}{\binom{52}{5}} $$