How many ways are there to shuffle a deck of cards if cards of the same value (but different suits) are indistinguishable from one another? For instance, the jack of hearts and the jack of spades are considered as the same card?
Initially, my intuition was to take $52!$ and subtract all duplicate shuffles (of which I believe there are $52 * 4!$). $52 * 4!$ comes from $4!$ different ways to arrange each card value, and $52$ different cards. Is this correct?
You can't just subtract, you need to divide. Think about this; for a given shuffle, how many other shuffles is it indistinguishable from? Well, you can permute the four aces in $4!$ ways to get another shuffle, which is indistinguishable since we are thinking of aces as identical. You can also simultaneously and independently shuffle the twos, threes, ... , queens and kings, each in $4!$ ways. Using the multiplication principle, the number of shuffles indistinguishable from this one is $(4!)^{13}$.
Now, imagine taking all $52!$ shuffles, and grouping them together so that all shuffles within each group are indistinguishable from each other. From the previous logic, each group has exactly $(4!)^{13}$ shuffles. Since $52!$ shuffles have been divided into equal groups of size $(4!)^{13}$, the number of groups must be $$ \frac{52!}{(4!)^{13}} $$
