Logic behind the combo of cards in a hand that contain only clubs In looking at a stats problem where you want all combos of a 5-card hand that contain at least one club, the approach I have is to find the combos of 5-card hands that do not contain clubs, and then subtract it from the total number of 5-card hand combos, so this would be the following:
$$\left(\begin{array}{c} 52 \\ 5 \end{array}\right) - \left(\begin{array}{c} 39 \\ 5 \end{array}\right)$$
I am pretty sure the above approach is correct. What I'm curious about is why the logic does not flow to the multiplication rule? Is it not that if we know all 5-card combinations of Hearts, Diamonds, and Spades, then we can multiply these together to get all possible 5-card combinations of these suits? By that logic, the following should be the same as the above, but its not.
$$\left(\begin{array}{c} 52 \\ 5 \end{array}\right)-\left(\begin{array}{c} 13 \\ 5 \end{array}\right)\cdot \left(\begin{array}{c} 13 \\ 5 \end{array}\right)\cdot \left(\begin{array}{c} 13 \\ 5 \end{array}\right)$$
Perhaps I'm not seeing something obvious. Can anyone shed any light?
 A: You said that you wanted to find 

all combos of a 5-card hand that contain only clubs

There are $13$ clubs to choose from, so this should be simply $$\binom{13}{5}$$
But then you calculated the number of hands that contain at least one club. Which do you want to find?
For example, a hand with 4 clubs and 1 spade would be counted in the second calculation, but not in the first.
A: Multiplying together the number of ways to do X, Y, and Z gives you the number of ways of doing X, Y, and Z together.  So your second equation represents the number of ways of taking $5$ hearts, $5$ diamonds, and $5$ spades, essentially forming a $15$ card hand.  I can't think of any meaningful description of the value you get when subtracting this number from the number of $5$-card hands, and it's certainly not the number of hands with at least one club.
If you added the values instead of multiplying, you would get the number of ways of taking $5$ hearts, $5$ diamonds, or $5$ spades.  So subtracting the sum from the number of $5$-card hands would give you the number of hands that are not all hearts, all spades, or all diamonds.  This would still not give you the number of hands with at least one club (which I'm assuming is what you actually want), however, as it would exclude the possibility of mixed-suit hands, eg $3$ hearts and $2$ spades.  
A: If you

find the combos of 5-card hands that do not contain clubs, and then subtract it from the total number of 5-card hand combos

then you find the combos with some clubs, not exactly five clubs. Right? ;)
In a deck of $52$ cards ($4$ suits, $52/4=13$), to find the combos of $5$ with exactly $5$ clubs, you must calculate
$$\binom{13}{5}= 1287.$$
Hence, the combos of $5$ with exactly $5$ cards of any suit is
$$4\cdot\binom{13}{5}= 4\cdot1287=5148.$$
