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

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Director
Joined: 18 Feb 2019
Posts: 581
Location: India
GMAT 1: 460 Q42 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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01 Mar 2019, 01:39
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4
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15% (low)

Question Stats:

78% (01:05) correct 22% (01:32) wrong based on 88 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: 107
GMAT 1: 570 Q39 V41
GMAT 2: 640 Q38 V31
GMAT 3: 650 Q42 V38
GMAT 4: 650 Q44 V36
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GPA: 3.75
Re: In how many ways can 4 students out of n students be selected to repre  [#permalink]

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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.

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

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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