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Problem#101: A student committee on academic integrity has

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Problem#101: A student committee on academic integrity has  [#permalink]

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New post 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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New post 21 Nov 2005, 06:18
10...x!/(2!(x-2)!)*(x-1)!/(1!(x-2)!)=90...solve for x...
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New post 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
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New post 21 Nov 2005, 06:37
combo(n,2)=90
n(n-1)/2=90
n^2-n-180=0
and no interger solution.
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New post 21 Nov 2005, 07:07
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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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New post Updated on: 21 Nov 2005, 07:30
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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.
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Re: ProjectGMAT- Last Problem in the book  [#permalink]

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New post 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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Re: ProjectGMAT- Last Problem in the book  [#permalink]

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New post 22 Nov 2005, 00:31
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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.
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New post 22 Nov 2005, 21:26
You guys are too good! Answer is 10.
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New post 22 Nov 2005, 21:35
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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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Re: Problem#101: A student committee on academic integrity has  [#permalink]

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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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