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by rijul007 » Wed Jan 25, 2012 10:31 pm
StoneBlack 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
no of ways of selecting 2 people from 8 = 8C2 = 28
no of ways of selecting 2 from remaining 6 = 6C2 = 15
no of waays of selecting 2 from 4 = 4C2 = 6
the remaining 2 will form a team

total no of selections = 28*15*6 = 28 * 90 = 2520

Option B
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by shankar.ashwin » Wed Jan 25, 2012 11:19 pm
The order in which the 4 teams are selected doesnt matter here..

So, its (8C2*6C2*4C2*2C2)/4! = 2520/24 = 105 E
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by LalaB » Wed Jan 25, 2012 11:23 pm
@rijul007 u should divide it by 4!


my approach 7*5*3*1=105

or use the formula (mn)!/(n!)^m*m!=(4/2)!/(2^4)*4!=7*5*3=105
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by StoneBlack » Wed Jan 25, 2012 11:28 pm
Thanks guys for replying.
The answer indeed is 105.

I cannot understand the logic applied to divide by 4!.Could you elaborate more on this?
Thanks.
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by shankar.ashwin » Wed Jan 25, 2012 11:41 pm
Hope this post clarifies
https://www.beatthegmat.com/doubles-t8119.html
StoneBlack wrote:Thanks guys for replying.
The answer indeed is 105.

I cannot understand the logic applied to divide by 4!.Could you elaborate more on this?
Thanks.
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by LalaB » Wed Jan 25, 2012 11:48 pm
StoneBlack wrote:Thanks guys for replying.
The answer indeed is 105.

I cannot understand the logic applied to divide by 4!.Could you elaborate more on this?
Thanks.
hm, will try to explain :)

for example , u have A B C D E F G H . grouping them in this way AB CD EF GH is the same as EF CD AB GH
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by GMATGuruNY » Thu Jan 26, 2012 3:45 am
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https://www.beatthegmat.com/a-group-of-8 ... tml#359208

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