# Combinations

• May 4th 2008, 08:41 PM
strgrl
Combinations
Hi guys!

Havings a problem with this... how many ways can you have 2 cards of one value and 3 other cards each of a different value? (assuming a standard 52 card deck)

My thought process was: C(13,1)xC(4,2)xC(12,3)xC(4,3)

Though that doesn't seem to be giving me the right answer. Any suggestions?

Thanks! :)
• May 5th 2008, 12:56 AM
Moo
Hello,

Quote:

Originally Posted by strgrl
Hi guys!

Havings a problem with this... how many ways can you have 2 cards of one value and 3 other cards each of a different value? (assuming a standard 52 card deck)

My thought process was: C(13,1)xC(4,2)xC(12,3)xC(4,3)

Though that doesn't seem to be giving me the right answer. Any suggestions?

Thanks! :)

1 card of a given value :
52 possibilities

Then remain 51 cards, including 3 of this value.
Another card of this value :
3 possibilities (because there are 3 left of this value).

There are 50 cards remaining. But we're only interested in 48 of them since we don't want to have the first value anymore.
1.
48 possibilities.

There remain 47 cards, including 3 of the value 1., which we are not interested in because we want them to be different.
2.
44 possibilities (47-3)

The same way, there remain 40 cards which do not contain the previous values.
3.
40 possibilities.

---> Number of possible combinations : 52x3x48x44x40

I hope my reasoning seems correct to you :x
• May 5th 2008, 09:28 AM
Soroban
Hello, strgrl!

I got a different answer . . .

Quote:

How many ways can you have 2 cards of one value and 3 other cards
each of a different value? (assuming a standard 52-card deck)

In Poker this is called "One Pair": a Pair of matching values and 3 Others.

There are 13 choices for the value of the Pair,
. . and ${4\choose2} = 6$ ways to get the Pair.
Hence, there are: . $13\cdot6 \:=\:{\color{blue}78}$ possible Pairs.

The three Others must be from the remaining 12 values.
. . There are: . ${12\choose3} \:=\:220$ choices of values.
For each card, there are: . ${4\choose1} = 4$ choices.
Hence, there are: . $220\cdot4^3 \:=\:{\color{blue}14,080}$ choices for the 3 Others.

Therefore, there are: . $78\cdot14,080 \:=\:\boxed{1,098,240}$ possible One Pair hands.