Four players choose 7 cards from a 52 card deck with the rest of the cards going back I'm doing a homework assignment (with answers uploaded from my TA) and have been stumped over this particular answer.
There are 4 players. Each player selects 7 cards from the deck. The remaining cards remain in the deck.
The answer is 52!/(7!7!7!7!24!)
What I'm stumped on is the 24!. Why is that included in the answer at all? If each player selects 7 cards from the deck, why is there a 24! at the end?
 A: If you think about it, the expression amounts to
$\binom{52}7\binom{45}7\binom{38}7\binom{31}7\binom{24}{24}$,
We could as well have left out the $\binom{24}{24}$ part in the above expression for the residue as it will always evaluate to $1$
But if you express it as $\frac{52!}{7!7!7!7!24!}$, the $24!$ in the denominator becomes mandatory, since by writing 52! in the numerator, you are compelled to distribute all 52 cards.
A: Imagine drawing 52 empty boxes on a large sheet of paper.  The first 7 will be the first player's cards.  The next 7 for the next player, etc.  And the final 24 slots are the leftovers.
Now shuffle the cards and fill all the slots.  There are 52! ways to do that.  But we don't care what order the first 7 are in.  So divide by 7!.  We don't care what order any of those five sets are in, so we divide by the number of ways each set can be arranged.  $\frac{52!}{7!7!7!7!24!}$
Probably over-explained that.  Just one way to think about it.  The thing to remember is that if you started with 52! in the numerator, then you need to address all 52 cards.  And since we don't care how those final 24 cards are arranged, we have to divide by 24!.
