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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
I have one basic question. For the problem above..
13c1 * 13c1 * 13c1 * 13c1 ... we need to have 2 cards.. when we choose 2 cards is the order in which we choose the cards is important?? like taking 2 and 5 is different from 5 and 2 ??? why is it 48 * 47 instead of 48c2 ??