jeeteshsingh wrote:

A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420

B. 2520

C. 168

D. 90

E. 105

We should divide this by 4! --> 2520/4!= 105, as the order of the teams does not matter.

You can think about this in another way.

For the first person we can pick a pair in 7 ways;

For the second one in 5 ways (as two are already chosen);

For the third one in 3 ways (as 4 people are already chosen);

For the fourth one there is only one left.

So we have 7*5*3*1=105

You can heck this:

combination-groups-and-that-stuff-85707.html#p642634There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

(4*8)!/(8!)^4*4! but the result was way off.

Appreciate the help.