Number of all possible different combinations of a non-standard card deck.

Question

Father and son play a game with a 52-card deck of cards. Each card of this deck has a number from 1 to 13 on it and there are 4 cards exactly the same for each number.

In how many different ways can these cards be placed in the deck?

My effort: I first began by considering the simpler problem where there are only 4 exact same aces, other than that the deck would be standard (2 colors, 4 symbols, 52 cards) except for the aces. To find all the possible different ways this specific deck can be arranged, I thought of considering the 4 aces as distinct ones, and then dividing by 4! since that's the number of different ways 4 items can be placed without repetitions.

Having solved the relaxed problem, I thought that by induction that the answer to the original question would be $\frac{52!}{(4!)^{13}}$.

Is my answer correct? Am I missing something big? Any help would be greatly appreciated.

Answer 1

To begin with, you can place the aces in $52 \choose 4$ ways. Then for each of those, you can place the twos in $48 \choose 4$ ways. And for each of those you have $44 \choose 4$ ways to place the threes. Expanding that out, we see that the aces, twos, and threes can end up arranged within the final deck in $$\begin{eqnarray} {52\choose 4}{48\choose 4}{44\choose 4} &=& \frac{52!}{(52-4)! 4!} \frac{48!}{(48-4)! 4!} \frac{44!}{(44-4)! 4!} \\ &=& \frac{52!}{(44-4)! (4!)^3} \end{eqnarray}$$ ways.

Proceeding in this way all the way up to the kings, for which there will always remain only ${4\choose 4}=1$ way to place them, we also note that the product will continue to telescope—each subsequent factor’s numerator cancels with its predecessor’s denominator. So the answer is, as you suggest $52!/(4!)^{13}$.

Answer 2

Yes, your answer is correct... The method I used included 52! as the total ways it could be arranged if we consider a standard deck and now I chose one type of (like an ace).. When we shuffle the 4 aces together since this being a non standard deck we would have considered a factor of 4! extra. Now there can be 13 such cases of choosing a card like an ace which raises 4! by a power of 13..

Hence the answer is 52!/{(4!)^13}