to arrange 8 people is 8!
Divided by 4 group of 2 people 4! x 2! x 2! x 2! x 2!

= (8 x 7x 6 x 5 x 4!) / (4! x 2! x 2! x 2! x 2!)
= 7 x 3 x 5
= 105

You have the following options of 8 teams divided into 4 teams of 2 persons:

-- -- -- --

1st -- = can be arranged in 8*7 ways and the order can be (AB or BA) so 2 ways of arranging them

2nd -- = can be arranged in 6*5 ways and the order can be (CD or DC) so 2 ways of arranging them

3rd -- = can be arranged in 4*3 ways and the order can be (EF or FE) so 2 ways of arranging them

4th -- = can be arranged in 2*1 way and the order can be (GH or HG) so 2 ways of arranging them

Also then 4 groups can be arranged in 4! ways like
ABCDEFGH
ABCDEFHG and etc in 4! ways

So total = 8!/(4!*2!*2!*2!*2!) = 105

-Deepak

Just another way I thought about the solution

Number of ways of selecting 2 people from 8 = 8C2
Number of ways of selecting 2 people from 6 = 6C2
Number of ways of selecting 2 people from 4 = 4C2
Number of ways of selecting 2 people from 2 = 2C2

Total number of way of selecting 4 teams = 8C2x6C2x4C2x2C2 --- (1)

Number of ways these 4 teams can be arranged is 4! ---- (2)

so total number of distinct ways = (1)/(2)
= 105

Doofus,
Good solution. I would have the same.

8c2*6c2*4c2*2c2/ 4! (since the order we pick the teams does not matter)

= 105

why shouldn't we multiply by 4!?

could someone explain why it matters how many different ways the 4 teams can be arranged?

It seems like all that matters is the distribution of the people among teams and not the order of the teams .. or the order of the people on the teams for that matter ..

Thanks for the help in advance.