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A student committee on academic integrity has 90 ways to select a pres

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A student committee on academic integrity has 90 ways to select a pres  [#permalink]

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New post 15 Feb 2016, 09:43
3
11
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

86% (01:22) correct 14% (01:52) wrong based on 246 sessions

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Re: A student committee on academic integrity has 90 ways to select a pres  [#permalink]

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New post 15 Feb 2016, 10:13
7
3
Bunuel wrote:
A student committee on academic integrity has 90 ways to select a president and vice president from a group of candidates. The same person cannot be both president and vice president. How many candidates are there?

A. 7
B. 8
C. 9
D. 10
E. 11

Let t = TOTAL number of candidates
So....
# of ways to select a president = x
# of ways to select a vice-president = x - 1
Total number of ways to select both = (x)(x - 1)
We're told that 90 ways are possible
So, (x)(x - 1) = 90
Expand: x² - x = 90
Rearrange: x² - x - 90 = 0
Factor: (x - 10)(x + 9) = 0
So, x = 10 OR x = -9
Since x cannot be negative, it must be the case that x = 10

Answer: D

Cheers,
Brent
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A student committee on academic integrity has 90 ways to select a pres  [#permalink]

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New post 15 Feb 2016, 10:07
2
Let the number of people to choose from = n
No of people to be chosen = 2
The number of ways 2 people can be chosen from n people=nC2
Now, the two people selected can be arranged in 2! ways

Hence nC2*2!=90 which gives n=10

So D it is
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Re: A student committee on academic integrity has 90 ways to select a pres  [#permalink]

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New post 16 Feb 2016, 09:48
xC1 * (x-1)C1=90
x^2 -x -90 =0
(X-10) (X+9) = 0
X= 10 , -9

-9 can't possible.

Hence 10 should be the answer
IMO D
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Re: A student committee on academic integrity has 90 ways to select a pres  [#permalink]

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New post 21 Feb 2016, 18:03
2
Bunuel wrote:
A student committee on academic integrity has 90 ways to select a president and vice president from a group of candidates. The same person cannot be both president and vice president. How many candidates are there?

A. 7
B. 8
C. 9
D. 10
E. 11



I think we can use permutation equation for this since order matters (first selection being president and second being vp).

You can test numbers in the permutation equation, and since the list of answers is in acending order, start with the middle value:

n = 9 --> 9! / (9-2)! = 56 too small

n= 10 --> 10! / (10-2)! = 90 --> right

answer: D
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Re: A student committee on academic integrity has 90 ways to select a pres  [#permalink]

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New post 01 Jun 2020, 06:18
BrentGMATPrepNow wrote:
Bunuel wrote:
A student committee on academic integrity has 90 ways to select a president and vice president from a group of candidates. The same person cannot be both president and vice president. How many candidates are there?

A. 7
B. 8
C. 9
D. 10
E. 11

Let t = TOTAL number of candidates
So....
# of ways to select a president = x
# of ways to select a vice-president = x - 1
Total number of ways to select both = (x)(x - 1)
We're told that 90 ways are possible
So, (x)(x - 1) = 90
Expand: x² - x = 90
Rearrange: x² - x - 90 = 0
Factor: (x - 10)(x + 9) = 0
So, x = 10 OR x = -9
Since x cannot be negative, it must be the case that x = 10

Answer: D

Cheers,
Brent


Hi Brent, in the solution above, you basically applied Slot method to solve the problem. As far as I know, in the Slot Method, the order is considered i.e. the following two cases are counted as separate
CASE 1: President is selected first, Vice-President is selected next
CASE 2: Vice-President is selected first, President is selected next

However, I do not understand here why the order of selection shall be considered since this problem is similar to selecting a team of two people from a group of people (For e.g. Selecting a team of 2 people from a group of 6 people can be done in C(6,2) = 6!/2! 4! = 15 ways ). Could you please enlighten me on this.
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Re: A student committee on academic integrity has 90 ways to select a pres   [#permalink] 01 Jun 2020, 06:18

A student committee on academic integrity has 90 ways to select a pres

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