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President/Vice President

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Senior Manager
Senior Manager
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Joined: 15 Feb 2011
Posts: 263
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Kudos [?]: 22 [0], given: 9

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President/Vice President [#permalink] New post 08 Aug 2011, 19:14
A student committee on academic ntegrity 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 students are in the group?
Manager
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Re: President/Vice President [#permalink] New post 08 Aug 2011, 19:27
let there be N students in the committee. To select president there are N choices. After selecting president, there are (N-1) members left. now to select V. president there (N-1) choices.

According to question N(N-1) =90

N^2 -N =90

N^2 -N - 90 =0

N^2 -10 N +9N -90 =0

(N-10)(N+9) =0, so N =10. so there are ten students in the committee
Senior Manager
Senior Manager
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Joined: 15 Feb 2011
Posts: 263
Followers: 4

Kudos [?]: 22 [0], given: 9

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Re: President/Vice President [#permalink] New post 08 Aug 2011, 20:07
Aj85 wrote:
let there be N students in the committee. To select president there are N choices. After selecting president, there are (N-1) members left. now to select V. president there (N-1) choices.

According to question N(N-1) =90

N^2 -N =90

N^2 -N - 90 =0

N^2 -10 N +9N -90 =0

(N-10)(N+9) =0, so N =10. so there are ten students in the committee


Sometimes i cant help but smile on my own thought process..anyhw..

Thanks AJ..Kudos to you...also, can u help me understand the below logic?

I came across a point in p&c where if you know how many subgroups of a certain size you can choose from an unknown original larger group, you can deduce the size of the larger group however, what i couldn't do is the actual calculation..

Can you help on how to solve the following equation..

The total number of women are x and if 2 more women are added for selection, exactly 56 different groups of 3 women could be selected..
Manager
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Re: President/Vice President [#permalink] New post 08 Aug 2011, 21:12
Well I have never solved such a problem, But I will try to solve it. you mean to say originally there were x number of women and then 2 women were added. then while choosing groups of 3 women from (x + 2) women there are 56 selections available. so what is x ?

so equation should be (x + 2)!/ 3! (x+2 -3)! = 56

we can do some cancelling and find (x +2)(x + 1) x/ 3! =56

take 3! =6 on right hand side, we get

(x + 2)(x +1) x = 56 * 6

Now we can see x, (x +1) and (x + 2) are consecutive numbers and 56 = 7 * 8

so basically (x + 2)(x +1) x = 8 * 7 * 6

so x= 6

Well this is the only method I can come up with :) let us wait for experts , they can give a better method than this and proper way of solving this question.
Re: President/Vice President   [#permalink] 08 Aug 2011, 21:12
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