Discrete Mathematics 8ed, R Johnsonbaugh, problem 6.5.28: "Find the probability of obtaining a bridge hand with 5–4–2–2 distribution, that is, five cards in one suit, four cards in another suit, and two cards in each of the other two suits."
Answer: $\frac{4C(13,5)\times3C(13,4)C(13,2)^2}{C(52,13)}$
I understand why there's a 4 in the numerator, it's because we can start picking 5 cards from 13 from any of the 4 suits. Then the 3 is there supposedly to say that, "ok now we can pick 4 cards form 13 from the other 3 suits", right?
But why is there not a "2" in the numerator, saying "and we can pick 2 cards from 13 from the third suit (of which we have 2 options)"? (In this case we'd finish the numerator simply by the factor $C(13,2)$ which would say "pick 2 cards from 13 in the last remaining suit, of which we have 1 option.")