If you are holding an ace and a king of different suits, how many ways are there to flop a straight?

My first approach was that there are 12 cards which could be the first card of the flop, 8 cards which could be the second card of the flop, and 4 cards which could be the final card of the flop. So, since each card is selected independent of the other cards, 12*8*4=384.

But when I did a little reading about combinations and permutations online, this didn't look like what I discovered. So, I tried a different approach. There are 3 sets of 4 cards each, Q,J&10. Since each Q can combine with each J, that is 4*4=16. And since that combination can combine with each 10, that is 16*4=64. That seemed a bit low compared to my first answer. Then I decided that while the order of the flop cards doesn't matter as concerns the final hand, a straight, it does matter as to the number of ways you can reach the straight. It does matter which set the first, second and third cards come from. Since the number of ways you can arrange the 3 sets is a permutation, here you would have 3! or 6 arrangements. So, I multiplied 64*6 and got 384 again. Encouraging, but everything said this was a combination problem and neither of these two methods looked like normal combination expressions of nCk or n!/(n-k)!*k!

So, I opened the function wizard on Excel and selected COMBIN. For the first flop card, I had to enter 12 and 1 since there were 12 acceptable cards but I was only choosing 1 of them. It just looked strange to write it as 12C1, but, it returned the answer 12. When I tried 8C1, it returned 8, and for 4C1 it returned 4. This looked like my first approach only written as 12C1*8C1*4C1*=384. At least this looked more "mathy."

Finally, I went back to my second approach of the 3 sets. But all I ended up with was 4C1*4C1*4C1*3!=384, although it too looked more "mathy."

Okay, so that exhausted what I know on the subject. It seems right, but I'm really not sure. Since, to find someone super knowledgeable in this area at 2+2, all you have to do is throw a stick, so to speak, I decided to post the question here in hope of some clarification.

Thanking you in advance.

First, second, third card is irrelevant, it's a 3-card flop. You must flop a Q,J,T and there are 64 combinations of those.

If you want the percentage, then total flop combinations is 19,600. Divide.

If for some reason you care about permutations (order matters), then it's simply 12*8*4= 384. But you shouldn't use this number for anything useful.

Using permutations is the same as using fractions which is the simplest way to do it:

12/50 * 8/49 * 4/48

=~ 0.327%


Using combinations it is

4*4*4 / C(50,3)

= 64 / 19600

=~ 0.327%

Each of those fours is actually C(4,1).