number of ways to obtain a pair Here is my attempt to find number of hands in a 5 cards poker game that gives one pair:

$ \binom{4}{2}\binom{48}{3} \cdot 13$ - 2pairs - full house
=1349088-123552-3744=1221792

But it is wrong. What did I do wrong? I know I have the right numbers for full house and two pairs. I dont need another solution for obtaining number of hands that holds one pair. I only need to get an explanation for why this approach to finding number of one pair is wrong.

My approach is to find any hand that has 1 pair and any other 3 cards and then subtract hands in that category that gives other poker combinations which should be full house and two pairs.
As far as I can see, your approach is:

Select a card value. Now, from the $4$ cards with that value in the deck, select $2$ of them to be in the hand. Next, any $3$ of the other $48$ cards in the deck can go in the hand.

Now since there are $13$ different values, multiply by $13$.

The hands which aren't pairs using this approach are 2 pair and full house, so subtract them off the final total.

Correct?



The problem seems to be that you're counting the hands with two pairs twice here.

Example: I select Ace. Now I have AA in hand. I select 553 as the other cards for the hand AA553.

But when I select 5, I can also select 55AA3 as a hand which is the same.



There aren't any further problems, so $1349088-2(123552)-3744=1098240$ is actually the correct answer.
You've counted all the hands that have a pair in, and then subtracted the hands with two pairs or a full house. You haven't subtracted the four-of-a-kind or triplet hands which also contain a pair.