Re: combination or permutation??
You're both almost right, but it's a bit trickier than this. If you use 38 then you're computing the probability of getting a straight without pairing any of the needed cards. But if you use 47, you'll double count those occurances (and more than double counts one with two pair or threes).
To do this correctly you start with 38, then you compute the probability of getting a straight and pairing one of the cards; getting a straight and pairing two of the cards; and getting a straight with three of a kind.
That's:
4*4*4*C(38,2) (no pairs)
3*4*4*6*38 (one pair)
3*4*6*6 (two pairs)
3*4*4*4 (three of a kind)
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