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group of x individuals

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group of x individuals

by alex.gellatly » Sat Jun 23, 2012 8:48 pm
How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed
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by sandeep_thaparianz » Sun Jun 24, 2012 4:20 am
OA should be D

Reframe the question and that is if we find X we can solve the question

From statement 1 we got exactly 126 teams with 5 members. Which is equal to 9C5= 9*8*7*5!/(5!*3!)=126 hence x is equal to 9


From statement 2 we got exactly 56 teams with 3 members. Which is equal to 8C3= 8*7*5!/(5!*3!)=56 hence x is equal to 8

So answer is D
Hope it helps
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by Anurag@Gurome » Sun Jun 24, 2012 5:07 am
alex.gellatly wrote:How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.
(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed
We need to determine the value of xC5.
Hence, we need to determine the value of x.

Statement 1: (x + 2)C5 = 126
So, (x + 2)(x + 1)(x)(x - 1)(x - 2)/(5!) = 126
--> (x + 2)(x + 1)(x)(x - 1)(x - 2) = 126*(5!)

Hence, product of 5 consecutive integers is 126*(5!) = (2*7*9)*(2*3*4*5)
Now, 126*(5!) can be written as the product of five integers in only one way : 5*6*7*8*9
Hence, (x + 2) = 9 --> x = 7

Sufficient

Statement 2: (x + 1)C3 = 56
So, (x + 1)(x)(x - 1)/(3!) = 56
--> (x + 1)(x)(x - 1) = 56*(3!)

Hence, product of 3 consecutive integers is 56*(3!) = (7*8)*(2*3)
Now, 56*(3!) can be written as the product of five integers in only one way : 6*7*8
Hence, (x + 1) = 8 --> x = 7

Sufficient

The correct answer is D.
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