How many hands (of size 5) of a cards from a standard deck with 4 of a kind are there? I was trying to count how many cards with 4 of a kind I could get from a standard deck of 52 cards in a hand of length 5.
i.e. 4 of the same value (rank) and a card with a different value.
I was told the answer was $13 \times 12 \times 4 = 624$ but I was a little confused why.
The explanation given to me was because a four of a kind can be completely specified by a sequence saying


*

*The rank of the four cards

*The rank of the extra card

*The suit of the extra card


However, there is a specific reason I am confused. For me this reasoning seems to be undercounting. Why do I think that, well, I believe that the suit of the cards with same rank should be important too because a 4 of a kind hand with {8 spades, 8 spades, 8 spades, 8 spades, Q hearts} is a different 4 of a kind than {8 clubs, 8 clubs, 8 clubs, 8 clubs, Q hearts}
i.e. does the suit of the cards with the same kind not matter?
Why is it not $13 \times 4 \times 12 \times 4$ i.e. 


*

*The rank of the four cards

*The suit of the four cards

*The rank of the extra card

*The suit of the extra card


I just don't understand why that is wrong.
Actually, that is wrong too, why is it not:


*

*The rank of the four cards

*The suit of each of the four cards in a combination way

*The rank of the extra card

*The suit of the extra card


As in why is the answer not:
$$ 13 \times C^4_1 \times C^4_1 \times C^4_1 \times C^4_1 \times 12 \times 4$$
or something that takes into account the different suits for the 4 of a kind cards?
 A: 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.
A: The four cards include all suits, and there is only one way to include all four suits.  It might be 8 hearts, 8 spades, 8 clubs, 8 diamonds, Q clubs.
A: To solve your confusion, we will not be able to create any four of a kind hands with {8 spades, 8 spades, 8 spades, Q hearts} or {8 clubs, 8 clubs, 8 clubs, Q hearts}. The reason for this is in a standard deck of 52 cards there are 4 suits with 13 distinct ranks in each suit. This means that once the 8 of clubs of Queen of hearts is chosen, we can no longer choose this card.
A solution to this problem can be broken down into two stages 
Stage 1: Choose the rank of the card for the 4 cards -> $\binom {13}{1}$ since there are 13 ranks
Stage 2: Choose the remaining card -> $\binom {48}{1}$ since we need one card and we chose 4 so there are only 48 left in the deck
Since the outcomes of each stage are independent of each other we can multiply the results of each stage to find the total number of hands with a four of a kind and one other card.
Total = $\binom {13}{1}$ x $\binom {48}{1}$ = 13 x 48 = 624
The other persons answer is the same thing in disguise, by choosing the remaining ranks and then suit you are simply choosing the remaining cards in the deck, notice that 12 x 4 = 48.
A: Here is another way to explain. 
In the case that "kind" means rank.
There are $13$ ranks in the $52$ cards. Each rank has exactly $4$ cards. To select $5$-card hand with $4$ cards of one rank:


*

*Firstly, select $4$ cards of one rank: $13$ ways to do this since there are $13$ or $\binom{13}{1}$ ways

*Secondly, select the last card: after selecting $4$ cards, we know have $48$ cards left. -> $48$ or $\binom{48}1$ or $12\times 4$ (remainedRanks*cardsPerRank) ways to select the last cards


Hence, there are $13 \times 48$ ways to select $5$-card hand of which $4$ cards are of one kind.
I hope this helps.
