Note first that in a standard 52-card deck, there is only one copy of each specific rank-suit card in the deck (i.e., only one Ace of spades, only one Ace of clubs, only one Ace of hearts, etc...)
When determining how many four-of-a-kind hands there are, you may choose to consider what suits the 4-tuple consists of as it helps to generalize the problem further. After the first step of picking what rank the 4-tuple is in, decide which 4 of the available 4 suits the cards will be, that is to say, there are $\binom{4}{4}$ "different" ways of having the suits selected for the 4-tuple. Due to the fact that $\binom{4}{4}=1$, many people choose to leave it out of the calculations because it is readily apparent to them that to have four of the same rank, you will necessarily use all four available suits and there exist no other configurations.
Down here instead, consider the related problem of asking to find how many full houses exist (again from a standard deck) (a full house consists of three cards of one rank and two cards of another)
Breaking it into steps:
1. Choose the rank of the triple: $\binom {13}{1} = 13 $
Choose the suits of the triple: $\binom {4}{3} = 4 $
Choose the rank of the double: $\binom{12}{1} = 12$
Choose the suits of the double : $\binom{4}{2} = 6$
Thus, the total number of full house hands in a standard deck is $13\cdot 4\cdot 12\cdot 6$
Note that in this case we can identify the triple from the double. If we asked how many two - pair there are you will need to divide by symmetry
Consider another similar situation in order to highlight the differences of using a nonstandard deck consisting of 65 cards: thirteen ranks available as usual (2,3,4,5,6,7,8,9,10,J,Q,K,A), but this time consisting of five different suits: (hearts, diamonds, clubs, spades, and swords)
How many four-of-a-kind hands exist for this nonstandard deck then?
Breaking it up into steps as before:
Choose the rank of the 4-tuple: $\binom{13}{1} = 13$
Choose the suits of the 4-tuple: $\binom{5}{4} = 5$
Choose the rank of the singleton: $\binom{12}{1} = 12$
Choose the suit of the singleton: $\binom{5}{1} = 5$
So, all together, there will be $13\cdot 5\cdot 12\cdot 5$ number of 4ofakind hands in this nonstandard deck.
Best Answer
Yes, that's right. You can't guarantee it with fewer, since if you only take $36$ cards you could get $9$ in each suit.