1. ## Card arrangement problem.

So the problem goes like this:

How many ways can a hand of five cards consisting of four cards from one suit and one card from another suit be drawn from a standard deck of cards?

So I was all like:

nCr(4, 13) * nCr(1, 13) because you can choose 4 from one suit of cards (which is 13 cards in total) and then 1 from the other...

I get 9295 but the book is like no: the answer is not 9295...

It is instead 111,540... So I'm all like ok:

Lets try nCr(4, 13) * nCr(1,39) basically saying it is every other suite (52-13), but that did not give the right answer.

And now I'm all stuck on this problem. Can you help me? Thanks in advanced.

2. Hello, thyrgle!

How many ways can a hand of five cards be drawn from a standard deck
consisting of four cards from one suit and one card from another suit?

We want a foursome and a singleton.

There are 4 choices for the suit of the foursome.
. . And there are: . $_{13}C_4 \:=\:\frac{13!}{4!\,9!} \:=$ 715 choices for the four cards.

There are 3 choices for the suit of the singleton.
. . And there are: . $_{13}C_1 \:=$ 13 choices for the singleton.

Therefore, there are: . $4\cdot715\cdot3\cdot13 \:=\:111,\!540$ possible hands.