Probability of Deck of Cards, ignoring suits

[Lenus]

I am looking for a way of calculating the possible combinations of a standard deck of 52 cards.

I am aware of the 52! number, which is the total no of combinations 52 cards can form in a deck, but would like to know how to determine the total no of combinations if the suits are ignored.

Thanks in advance.

 

[Doug Huffman]

I do not see an expression for suits in 52!

 

[Stephen Tashi]

the possible combinations of a standard deck of 52 cards.

Do you mean "permutations" instead of "combinations"?

 

[Lenus]

I do not see an expression for suits in 52!

??
isnt the number of ways of arranging 52-card deck 52! ?

 

[Doug Huffman]

The cards could as well be of 52 unique words, still with no suits.

 

[Lenus]

Do you mean "permutations" instead of "combinations"?

Yes, in fact I do.

All I need is the way of calculating the number of ways of arranging the deck of cards without suits, i.e. Qh As 7d would be same as Qs Ad 7c.

 

[Lenus]

The cards could as well be of 52 unique words, still with no suits.

Sure, it would be the same problem, although we need to have 13 unique words, 4 each in such a deck of 52 cards to arrive to my initial question again.

 

[mathman]

Wild guess: $\frac{52!}{4!^{13}}$

 

[Lenus]

mathman,

thanks and it looks like the right approach but would appreciate a little explanation, please.
As an example, how would that formula be transformed for the deck of 8 cards with 4 aces and 4 kings?

would that be 8!/4!^2?

 

[mathman]

mathman,

thanks and it looks like the right approach but would appreciate a little explanation, please.
As an example, how would that formula be transformed for the deck of 8 cards with 4 aces and 4 kings?

would that be 8!/4!^2?

Yes. Numerator is all permutations of deck. Each term of denominator is permutation of cards of the same rank.

 

[Lenus]

I have manually double-checked the result from the formula for the deck of 8 and it is just spot on!

Appreciate your help, thanks a lot!