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Re: A student committee on academic integrity has 90 ways to select a pres [#permalink]
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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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]
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]
Even though you got to the answer, the approach probably is not correct. There is no mention of arrangement in the question. Can you please clarify why you did the arrangement?


rguglani wrote:
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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A student committee on academic integrity has 90 ways to select a pres [#permalink]
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I do not use explanations because you can copy my text. I think you should be familiar with the rules of academic integrity. This is a necessary article https://worldfinancialreview.com/academic-integrity-is-important-for-students/ for students or schoolchildren who like to solve tasks not honestly.
x = x - 1
(x) (x - 1)
(x) (x - 1) = 90
x² - x = 90
x² - x - 90 = 0
(x - 10) (x + 9) = 0
x = 10 / x = -9
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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Re: A student committee on academic integrity has 90 ways to select a pres [#permalink]
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