P & C

I think similar questions have been posted before

Anyways I can use the practice.

There are 8 ppl. so there are 8! ways to arrange then.

However you want to choose four teams each with 2 ppl in them. Each team can be arranged 2! ways.

so the answer is 8!/(2!*2!*2!*2!) = 2520

You can do it an alternative way as well.

You have 8 ppl. You want to choose 2. So first team = 8C2,

out of 6 ppl remaining you want to choose 2 more... so 6C2,

and so on.. you get 4C2 and 2C2 for the remaining teams.

Multiply the lot you get 8C2*6C2*4C2*2C2

You have to further divide 2520/4! as the teams are of the same size and are interchangeable(order does not matter) = 105.

https://www.beatthegmat.com/combinations-t44119.html