How many 5 card hands from a 52 card deck have 4 cards of 1 suit and 1 card of another? When solving this problem using n-choose-k, it seems there is more than one way to answer this question:
Why is "choose(4,1) * choose(13,4) * choose(39,1)" = 111540, the correct answer, while "choose(4,2) * choose(13,4) * choose(13,1)" = 55770 is incorrect?
 A: 
Why is "$\dbinom 4 1\cdot \dbinom {13} 4 \cdot \dbinom {39}1  = 111540$", the correct answer

This counts ways to choose 1 from 4 suits, 4 from 13 cards in that suit, and 1 from 39 the remaining cards.

while "$\dbinom 4 2 \cdot \dbinom {13}4\cdot \dbinom {13}1 = 55770$" is incorrect?

You counted ways to select two from four suits, 4 from 13 cards in one suit, and 1 from 13 cards in the other.   However you neglected to select which suit was for which; hence you only have half the required value.
$$\dbinom 4 1\cdot \dbinom {13} 4 \cdot \dbinom {39}1 ~{= \dbinom 4 1\cdot \dbinom {13} 4 \cdot \dbinom 31 \cdot \dbinom {13}1 \\= \dbinom 4 {1,1,2}\cdot \dbinom {13} 4 \cdot \dbinom {13}1\\= \dbinom 4 2\cdot \binom 2 1 \cdot \dbinom {13}4\cdot \dbinom {13}1}$$
A: The 2nd solution does not consider all remaining suits for the 5th card, resulting in too few combinations.
Imagine we change to a deck with only rank 1-4 and 3 suits (A, B, & C) = 12 cards.
Given that and also noting that:
   choose(4,1) * choose(13,4) * choose(39,1), can also be written as
   choose(4,1) * choose(13,4) * choose(13,1) * choose(3,1), the above question becomes:
Why is "choose(3,1) * choose(4,4) * choose(4,1) * choose(2,1)" correct vs
         "choose(3,2) * choose(4,4) * choose(4,1)" ?

choose(3,1) * choose(4,4) * choose(4,1) * choose(2,1) = 24
A1,A2,A3,A4 with each of the remaining card B1-B4,C1-C4 (8 combinations) plus
B1,B2,B3,B4 with each of the remaining card A1-A4,C1-C4 (8 combinations) plus
C1,C2,C3,C4 with each of the remaining card A1-A4,B1-B4 (8 combinations) for a total of 24

Compare this to:

choose(3,2) * choose(4,4) * choose(4,1) = 12
AB:  A1,A2,A3,A4 with each of B1-B4 (4 combinations) plus 
BC:  B1,B2,B3,B4 with each of C1-C4 (4 combinations) plus
CA:  C1,C2,C3,C4 with each of A1-A4 (4 combinations) for a total of 12

We are missing 12 combinations!
In the first row, where are A1,A2,A3,A4 with each of C1-C4?  The same goes for the 2nd and 3rd rows.
Using choose(3,2),  is using only the cards of 1 other suit for the last card, where
using choose(3,1) * choose(2,1) is using each of the remaining suits for the the last card.
