Hi,

Can you please post the OA as well? Sometimes the OA help a great deal in solving the question

chaitanyareddy wrote:Hi Please let me know what is the issue in my approach regarding solving the following problem.


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?

My approach :

No. of ways for selecting team A : 8C1*7C1.
No. of ways for selecting team B : 6C1*5C1.
No. of ways for selecting team C : 4C1*3C1.
No. of ways for selecting team D : 2C1*1C1.

so total 100.

You have the correct starting point, but what you are missing here is that the order of the people within each team does not matter. Say you have x and y in Team A, using your method xy is not the same as yx.

So, you need to divide each of those by 2 to make the order irrelevant

so, this should give you 8x7/2 x 6x5/2 x 4x3/2 x 2x1/2 = 56/2 x 30/2 x 12/2 x 1 = 28 x 15x6x1 = 28 x 90 number of ways of forming 4 teams of 2 each.

Now, furthermore, the order of the teams themselves shouldn't matter. If your teams are ABCD, then its the same as BCAD etc., so you divide futher by 4!

28 x 90 / 4 x 3 x 2 = 105

Can you check if that is the OA?

OA 105

very good explanation kmittal82