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.