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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