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# Problem#101: A student committee on academic integrity has

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GMAT Club Legend
Joined: 29 Jan 2005
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21 Nov 2005, 05:32
Problem#101:

A student committee on academic integrity has 90 ways to select a President and VP from a group of candidates. The same person cannot hold both positions. How many students are in the group?

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VP
Joined: 30 Sep 2004
Posts: 1434
Location: Germany

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21 Nov 2005, 06:18
10...x!/(2!(x-2)!)*(x-1)!/(1!(x-2)!)=90...solve for x...
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Director
Joined: 13 Nov 2003
Posts: 771
Location: BULGARIA

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21 Nov 2005, 06:23
If we select first in 10, second in 9 ways this makes 90 ways in total . Seems like 10 students
Intern
Joined: 07 Sep 2005
Posts: 15

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21 Nov 2005, 06:37
combo(n,2)=90
n(n-1)/2=90
n^2-n-180=0
and no interger solution.
SVP
Joined: 28 May 2005
Posts: 1589
Location: Dhaka

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21 Nov 2005, 07:07
1
we know that we are going to select 2 candidates President and VP.

there are 90 ways to to select them. because one is pesident and one is Vp so order matters and it should be permutation.

we can create an equation n P 2 =90

or n! /(n-2)! = 90

or n(n-1)(n-2)! /(n-2) ! =90

or n^2 -n-90 =0

or (n-10)(n+9) =0

so n = 10 becasue n can not be negative 9

so there are 10 students in the group.
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Senior Manager
Joined: 27 Jun 2005
Posts: 479
Location: MS

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Updated on: 21 Nov 2005, 07:30
2
10
let assume there are n student

so no of ways a president can be selected = nC1

now, the number of ways VP can be selected = (n-1)C1

total no of ways = nC1*(n-1)C1

90=nC1*(n-1)C1
n(n-1) = 90

hence n = 10

Originally posted by cool_jonny009 on 21 Nov 2005, 07:28.
Last edited by cool_jonny009 on 21 Nov 2005, 07:30, edited 1 time in total.
VP
Joined: 22 Aug 2005
Posts: 1075
Location: CA
Re: ProjectGMAT- Last Problem in the book  [#permalink]

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21 Nov 2005, 07:29
GMATT73 wrote:
Problem#101:

A student committee on academic integrity has 90 ways to select a President and VP from a group of candidates. The same person cannot hold both positions. How many students are in the group?

nC1 * n-1C1 = 90
n*(n-1) = 90
n = 10
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Senior Manager
Joined: 14 Apr 2005
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Re: ProjectGMAT- Last Problem in the book  [#permalink]

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22 Nov 2005, 00:31
2
GMATT73 wrote:
Problem#101:

A student committee on academic integrity has 90 ways to select a President and VP from a group of candidates. The same person cannot hold both positions. How many students are in the group?

This is a permutation question as the order is important. Let their be "N" students. We are given that NP2 = 90
i.e n!/(n-1)! = 90
i.e n(n-1) = 90
n^2-n-90 = 0
Solving this quadriatic equn we get n = 10 or n=-9, n = -9 is not possible. Hence n = 10. or 10 students in the group.
GMAT Club Legend
Joined: 29 Jan 2005
Posts: 5000

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22 Nov 2005, 21:26
You guys are too good! Answer is 10.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 4851
Location: Singapore

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22 Nov 2005, 21:35
2
1
Permutatino should be used. Consider two people, Bill and Charlie. Bill being preseident and Charlie being VP is different from Biill being VP and Charlie being president.

So XP2 = x!/(x-2)! = 90 --> x is the number of people

x!/(x-2)! = 90
x*(x-1)*(x-2)!/(x-2)! = 90
x(x-1) = 90
x^2 - x - 90 = 0

x=10
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25 Sep 2017, 10:01
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Re: Problem#101: A student committee on academic integrity has &nbs [#permalink] 25 Sep 2017, 10:01
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