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In how many ways can 4 students out of n students be selected to repre

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GMAT 1: 490 Q47 V13
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In how many ways can 4 students out of n students be selected to repre  [#permalink]

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New post 01 Mar 2019, 01:39
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84% (01:05) correct 16% (02:16) wrong based on 31 sessions

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In how many ways can 4 students out of n students be selected to represent the school in a competition?

I.If there were 3 more students, number of ways of selecting 4 students would be 840.
II. If there were 3 less students, number of ways of selecting 4 students would be 360.
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Joined: 16 Oct 2011
Posts: 82
GMAT 1: 620 Q37 V38
GMAT 2: 640 Q36 V41
GMAT 3: 650 Q42 V38
GMAT 4: 650 Q44 V36
GMAT 5: 570 Q31 V38
Re: In how many ways can 4 students out of n students be selected to repre  [#permalink]

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New post 01 Mar 2019, 11:07
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kiran120680 wrote:
In how many ways can 4 students out of n students be selected to represent the school in a competition?

I.If there were 3 more students, number of ways of selecting 4 students would be 840.
II. If there were 3 less students, number of ways of selecting 4 students would be 360.



IF YOU FIND MY SOLUTION HELPFUL, PLEASE GIVE ME KUDOS

There are three pieces of information that are helpful in solving this problem. The first is that we can solve any equation in one variable (also, we can solve for n variables as long as we have n equations). The second, is that the number of ways that k things can be choses from a larger group of n things is n!/((k!)(n-k)!)
The third piece of information that is helpful, is that n! can be written as n(n-1)(n-2)(n-3).....(n-a)! where a is some integer

(1) If there were 3 more students, the number of selecting 4 students would be 840. Using the formula, we have n+3 C 4 = (n+3)!/((4!)(n+3 - 4)! = 840
multiplying by 4!, gives (n+3)!/(n-1)! = 840*4!. We can then rewrite the LHS as (n+3)(n+2)(n+1)(n)(n-1)!/(n-1)! = (n+3)(n+2)(n+1)(n) =84-*4!. We can solve this equation using some appropriate mathematical technique - sufficient

(2) If there were 3 less students, the number of ways of selecting 4 students would be 360. Then we have n-3 C 4 = (n-3)!/((4!)(n-3-4)!) = 360.
Then (n-3)!/(n-7)! = 4!360 and (n-3)(n-4)(n-5)(n-6)(n-7)!/(n-7)! = 4!*360, and (n-3)(n-4)(n-5)(n-6) = 4!*360 Sufficient

The answer is D
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Re: In how many ways can 4 students out of n students be selected to repre   [#permalink] 01 Mar 2019, 11:07
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