saurabhgoel 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

Please help with the explanation of the solution !!

\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2, ...

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 check similar problems:

probability-88685.html?hilit=different%20items%20divided%20equallyprobability-85993.html?highlight=divide+groupscombination-55369.html#p690842sub-committee-86346.html?highlight=divide+groupsThere 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}Hope it helps.

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