<s>[INST] Probability that exactly three cards in at least one of the suits in an eight card hand from a standard deck. so my solution was to...
a) find the number of hands where exactly one suit has exactly 3 cards
b) find the number of hands where exactly two suits have exactly 3 cards
c) add a) and b) then divide by c(52,8). 
so starting with b) because it's slightly easier...
c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit...
c(3,1) way to pick rest of suits and c(13,3) ways to pick cards in that suit...
2 cards left among 26, so just c(26,2).
so answer for b) would be - > 12*c(13,3)^2*c(26,2)
then going with a)
c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit...
c(39,5) ways to pick rest of cards but must consider that we want to exclude possibility of having another 3 in another deck...
so what i did was count the ways in which we could have 3 cards in a suit among the 3 suits and 5 cards we need...
so c(3,1) ways of choosing a suit along with c(13,3) ways of selecting 3, and c(26,2) ways of selecting rest of the suits...
so c(39,5) - 3*c(13,3)*c(26,2) is my answer for a). 
add them and divide by c(52,8) but i get a probability that is way too high...
 [/INST] The numerator:
${13\choose 3}{13\choose 5}{13\choose 0}{13\choose 0}{4\choose 1,1,2} + {13\choose 3}{13\choose 4}{13\choose 1}{13\choose 0}{4\choose 1,1,1,1} + {13\choose 3}{13\choose 3}{13\choose 2}{13\choose 0}{4\choose 2,1,1,1}  + {13\choose 3}{13\choose 3}{13\choose 1}{13\choose 1}{4\choose 2,2}+{13\choose 3}{13\choose 2}{13\choose 2}{13\choose 1}{4\choose 1,2,1} = 499163808$
$\frac {499163808}{{52\choose 8}} = 66.3\%$
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