How to count the combinations to get a pair in the hole.

[King_J]

I know that there are six combinations of AA and I know that there are 16 combinations of AK. To figure that out I just take 4*4=16. If one king is gone I just take 4*3=12 and find out that there is 12 combinations left.

But I just cant figure out how to count the combinations of for an example AA. Plz help me with this. I am going craaaazy!

Thx

[BruceZ]

[ QUOTE ]
I know that there are six combinations of AA and I know that there are 16 combinations of AK. To figure that out I just take 4*4=16. If one king is gone I just take 4*3=12 and find out that there is 12 combinations left.

But I just cant figure out how to count the combinations of for an example AA. Plz help me with this. I am going craaaazy!

Thx

[/ QUOTE ]

C(4,2) = 4*3/2 = 6. The divide by 2 is so we don't count each combination twice.

[Precision1C]

If one A is out the number of combinations of AA is C(3,2)=3!/(2!1!)=3. C(n,k)=number of combinations of k items from set of size n= n!/(k!*(n-k)!)

[Gator]

SH SC SD HC HD CD -- 3 factorial -- if one of a suit is missing, i.e. hearts, that leaves 3 (in this example SC, SD, CD) where S= spades, H = hearts, etc.

[Eldog605]

You want to know why there are 6 combos for aces? Simple. The Ac has three partners...The Ah, As, Ad. That's three combos right there. Next, we take the Ah...we already counted Ah/Ac, so Ah only has 2 combos...As and Ad. 2 + 3 = 5. Next, we take the As...its already been crossed with the Ac and the Ah, so all we have left is As and Ad. And by now the diamond has gotten around, so there are no more combos to make. 3+2+1=6.
1. Black Aces
2. Red Aces
3. Heart/Club
4. Heart/Spade
5. Diamond/Club
6. Diamond/Spade