# Math Help - Combinations - basics

1. ## Combinations - basics

Hello!

So I am just starting out with combinations and I'm a little confused and could very much use some advice. I'm working on a problem, and I keep running in circles, so let me start from there.

The problem is:
"How many hands in a 5-card poker hand can have exactly 1 pair (2 of the same kind) in a 52 card deck?"

This is what I've tried to do:
Order doesn't matter, so I figure it's combinations.

To have a pair, I need one card from a suit of 13 cards, and another card from a different suit of 13 cards. There are 4 suits in a deck.

So I took one card from the first suit C(13,1) and then add it to one card from another suit C(13,1) and add them to the other 3 cards in a hand, C(52,3)?

And the final answer would be...
C(13,1) + C(13,1) + C(52,3)

I am not sure if this is in any way correct and I am worried I'm going to learn this completely the wrong way. Any advice would be great! Thank you.

2. There are 13 different kinds of cards, so the total number of

combinations possible if we choose 2 cards is

$13C(4,2)=78$

There are 48 possible choices for the 3rd card, 44 possible

choices for the 4th card and 40 possible choices for the

5th card, but the last 3 cards can be chosen in any

order so we have to divide by the number of possible

arrangements for 3 cards, 3!=6. So, we have

$\frac{13C(4,2)(48)(44)(40)}{6}=1098240$

3. Hi Galactus and thanks for helping me with this.

I have a question though on the last part, where you determine the 3rd, 4th and 5th cards.

I don't understand this part - if I have have selected 2 cards already, shouldn't there be 50 left for the 3rd card, and then when that is selected, 49 for the 4th and 48 for the 5th? I understand that 48, 44 and 40 are there because we have 4 suits - is this done to avoid having 3, 4 or 5 of a kind?

I'm a little confused on this, thanks again!

4. Hello, mander!

Galactus is absolutely correct . . .

I don't understand this part -
if I have have selected 2 cards already, shouldn't there be 50 left for the 3rd card,
and then when that is selected, 49 for the 4th, and 48 for the 5th?

Is this done to avoid having 3, 4 or 5 of a kind? . . . . yes

They asked for exactly one pair.

So we want two cards with the same value
. . and the other three must not match the pair or each other.

There are 13 choices for the value of the Pair.

Suppose they are Aces.
Then there are: . ${4\choose2} \:=\:6$ ways to get a pair of Aces.

There are indeed 50 cards left in the deck,
. . but we do NOT want to draw another Ace, right?
So we have only 48 choices for the third card.
. . Suppose it is a Jack.

The fourth card must NOT be another Ace nor another Jack.
. . So there are only 44 choices for the fourth card.

Get the idea?

5. OH!
I think I understand it now!

Ok, so I'm going to practice a couple more, but that seriously helped me a ton. Thanks Galactus and Soroban!