Two distinct packs of 52 playing cards are shuffled together

104 choices for the first card;
102 choices for the second card, as in 103 cards left (after we chose the first card), there will be 102 different cards from the chosen one;
100 choices for the third card, 102 left and 102-2 are different from two cards already chosen;
...
54 choices for the 26th card.

54*56*...*104 possibilities. BUT we must divide this number by 26! as the order doesn't matter here.

So total # of ways is: \(\frac{54*56*...*104}{26!}=\frac{2^{26}*27*28*...*52}{26!}=\frac{2^{26}*52!}{26!*26!}\)

OR:
If we choose 26 cards from one deck, clearly there won't be two cards of the same suit and same denomination. Now any card chosen can be either from the first deck or from the second, so any card has two options.

Choosing 26 from 52: \(C^{26}_{52}\);
Any card from these 26 has two options 2*2*...*2 - 26 times: \(2^{26}\)

So total # of ways is: \(2^{26}*C^{26}_{52}\) (this is the same as above).

P.S. I doubt this is GMAT question.