Card Counting Problem I tried these homework problems but think I ended up double counting on a few. I will post the question followed by my answer. I need help checking that they are correct or what the answer really is. * * * **Consider a game where 4 cards are dealt from a standard 52 card deck. Determine the number of hands that contain:** **(a) 4 different suits** $13C1^4 \cdot 4C1 \cdot 3C1 \cdot 2C1 \cdot 1C1 = 685,464$ **(b) 4 different suits and 4 different weights** $13C1 \cdot 4C1 \cdot 12C1 \cdot 3C1 \cdot 11C1 \cdot 2C1 \cdot 10C1 \cdot 1C1 = 411,840$ **(c) 4 consecutive ranks** $11C1 \cdot 4C1^4 = 2816$

If we consieder hands as unordered sets of cards (i.e. there are ${52\choose 4}$ hands), then there are $13^4$ hands with four different suits ($13$ choices for each suit), $4!{13\choose 4}=\frac{13!}{9!}$ hands with four different suits and four different weights (choose four weigts and permute them to the suits), $(13-3)\cdot 4^4$ hands with consecutive ranks (choose wone of $10$ "starting points" and assign suits).

If you consider hands as ordered sets instead (so that there are $\frac{52!}{48!}$ hands in total), each of the above results must be multiplied with $4!$.