ashesh.rajhans wrote: In how many ways can one choose 6 cards from a normal deck of cards so as to have all
suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6
Case 1: 3 cards of one suit, the other 3 cards of the remaining 3 suits
Number of suit options for the 3 cards in red = 4. (Any of the 4 suits.)
Number of ways to choose 3 cards of this suit = 13C3 = (13*12*11)/(3*2*1) = 13*2*11.
Number of card options for the second suit = 13.
Number of card options for the third suit = 13.
Number of card options for the fourth suit = 13.
To combine these options, we multiply:
4*13*2*11*13*13*13 = 13�(4*2*11) = 13�(88).
Case 2: a pair of one suit, a pair of another suit, the other 2 cards of the two remaining suits
Number of ways to choose two suits for the two pairs in red = 4C2 = (4*3)/(2*1) = 6. (Any 2 of the 4 suits.)
Number of ways to choose 2 cards for the first pair in red = (13*12)/(2*1) = 13*6.
Number of ways to choose 2 cards for the second pair in red = (13*12)/(2*1) = 13*6.
Number of card options for the third suit = 13.
Number of card options for the fourth suit = 13.
To combine these options, we multiply:
6*13*6*13*6*13*13 = 13�(6*6*6) = 13�(216).
Total ways = Case 1 + Case 2 = 13�(88) + 13�(216) = 13�(88 + 216) = [spoiler]13�(304[/spoiler]).
None of the answer choices is correct.
Ignore this problem, which is far too complex for the GMAT.
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