Permutation with standard deck of cards

Question

Given a standard deck of cards, there $52!$ are different permutations of the cards. Given two identical standard decks of cards, how many different permutations are there?

Comments

@Soumya Tiwari 

can you explain in more details?

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So suppose you have 6 cards a,b,c &a,b,c so total ways of arranginging is 6! Now as every pair identical (a,a) (b,b) (c,c) has been counted twice so we need to divide by 2! For each identical cards

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Soumya Tiwari ok ok got your point

Answer 1

Permutation of n objects with m repeated elements = $\frac{n!}{m!}$

Why? Suppose we have to permute AAB. Then 3!=6, i.e. Let first A be 1, second A be 2 and B be 3. Then following permutation are possible:

123 (AAB)

132 (ABA)

213 (AAB)

231 (ABA)

312 (BAA)

321.(BAA)

But in our case we see that 123=213, also 132=231 and similarly 312=321.

These repetitions are because of  2 repeated entries of A, wherein each gives 2! similar permutations. And therefore we need to divide by 2!.

So in our original question, we have 2 repeated entries of each of the 52 cards. For example, 3 of hearts of Deck-1 and Deck-2. SO total n=104, and m=2 but 52 different repeatitions.

Therefore $\frac{104!}{2!2!2!...52\: times} = \frac{104!}{2^{52}}$