Leo2008 wrote:My Solution to this problem is different from what the textbook gives but I am very sure of my answer....and I think the book solved it the wrong way.


Two packs of 52 playing cards are shuffled together. Find the number of ways in which a man can be dealt 26 cards such that he does not get two cards of the same suit and same denomination ?

My answer is (2^26)(56!/26!)

the book answer is (2^26)(56C26)

Any explanation or thoughts ?

NOTE: i'll assume that you meant to type 52 in both places where you actually typed 56. if you really do mean 56, then i have no idea where you got that.

for reasons that i can't really pinpoint, this problem strikes me as unlike the authentic gmat probability problems.

nevertheless, on to the solution:

(actually i shouldn't say 'the' solution; more like 'one possible solution')

consider the number of options available for each card:
first card: 104 possibilities
second card: 102 possibilities (the first card is gone, and its counterpart from the other deck is ineligible to be selected)
third card: 100 possibilities (the first and second cards, as well as their counterparts, are gone)
fourth card: 98 possibilities
...
twenty-sixth card: 54 possibilities

so that's 104 x 102 x 100 x ... x 54
= (2x52) x (2x51) x (2x50) x ... x (2x27)
= (2^26) x (52x51x50x...x27)
= (2^26)(52!/26!)

this is probably where your solution stopped.

BUT

here's the problem: you have to account for redundancy. if you just multiply the possibilities, you have a permutation: i.e., if you get the same 26 cards in a different order, then you will (mistakenly) count that as 'different'. you don't want that to happen, so you have to divide by 26!, which is the number of different rearrangements in which each group of 26 cards will actually appear.

so you get
(2^26)(52! / 26!26!)
...and that's identical to the book solution.

btw, what book?