I’m learning about combinatorics and need help. I understand that there are 16 possible combinations of AK but I’m not mathing right.

Using the formula nCr=n!/(r! * (n-r)!)

I am getting confused. To figure out the possible AK combos, wouldn’t the answer be 8C2 or 4C1 ? Of course both of those are wrong.

Or is the answer 8C2-4C1?

So here’s how I’m going about things.

For the first card you have a total of 8 items, and of them you are selecting 1. Therefore 8C1. For the second card you have a total of 4 cards and you are choosing 1. Therefore 4C1. The sum total of 8C1 and 4C1 is the number of combos for 2 specific but different ranks of cards. But that only gives 12 combos. Do I have to add another 4C1 because it is not known if the first card is an ace or a king?

Looking for help.

@desertdog
@artymcfly

Apologies if I'm misunderstanding the question.

If you are asking how many ways there are in a HE hand to make AK (or any non paired hand) the answer is 4*4 = 16

No advanced complex formula needed.

Again, if that's not what you're asking, I apologize.

Yes I’m wondering how to derive the number of combinations of AK using combinatorics formulas.

What you helped with is that the answer is the number of combos for the first card times the number of combos of the second card.

Essentially 4C1 * 4C1

So for Omaha, if you were trying to determine the number of combinations for you hole cards for AAKK, it would be 4C1 * 3C1 * 4C1 * 3C1 right?

Why?

Why use an advanced calculation when 4th grade arithmetic works perfectly?

Because it is the foundation for advanced calculations.

No, I think it would be 4C2*4C2.

Makes sense, thank you, welt.