So to make this clear, this is not just n choose 2. what im talking about is related to this: Combination of splitting elements into pairs but also all pairs less than 6.
so basically for 3 values 1,2,3 you have 3 pairings: $${(1,2),3}, $$ $${(2,3),1},$$ $${(1,3),2}$$ for 4 values you have all single pairs and 2 left over:
$${(1,2),3,4},$$ $${(1,3),2,4},$$ $${(1,4),2,3},$$ $${(2,3),1,4},$$ $${(2,4),1,3},$$ $${(3,4),1,2}$$ all double pairings: $${(1,2),(3,4)},$$ $${(1,3),(2,4)},$$ $${(1,4),(2,3)}$$
so a total of 6+3 = 9 pairings.
for 5 there is: 1 pair,3 left over and 2 pairs, 1 leftover and so on.
I'm wondering if there is a closed formula for this and I am looking for the value of 13.
I would really appreciate the help.
You can get a formula (although maybe not very closed-form, you would definitely want a computer to do this for you) via that logic. First with one pairing, you have $n \choose 2$ possibilities. For 2 pairs, you have ${n\choose2} {n-2\choose 2}$, but notice that this has an order given to the two pairs, so we divide by 2!. For 3 pairs, you get $\frac1{3!}{n\choose2}{n-2\choose2}{n-4\choose2}$, and so on. So, you can write the answer as a sum.
From a quick calculation, you can find the values here: https://oeis.org/A001189 So for 13, there are 568503 pairings.