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There are n members in a certain department, including Michael. Two re

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Bunuel
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There are n members in a certain department, including Michael. Two re [#permalink] New post   10 May 2021, 08:45
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There are n members in a certain department, including Michael. Two representatives are to be selected to attend a company conference. If there are 55 possible combinations in which Michael is not selected, what is the value of n?

A. 11
B. 12
C. 15
D. 18
E. 22
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Bunuel
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Re: There are n members in a certain department, including Michael. Two re [#permalink] New post   16 May 2021, 02:00
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Bunuel wrote:
There are n members in a certain department, including Michael. Two representatives are to be selected to attend a company conference. If there are 55 possible combinations in which Michael is not selected, what is the value of n?

A. 11
B. 12
C. 15
D. 18
E. 22


We are basically told that we can select 55 different groups of two out of n-1 people (without Michael), thus given that \(C^2_{n-1}=55\) --> \(\frac{(n-1)!}{(n-1-2)!2!}=55\) --> \(\frac{(n-1)!}{(n-3)!2!}=55\) --> \(\frac{(n-2)(n-1)}{2}=55\) --> \((n-2)(n-1)=110\) --> \(n=12\) (by trial and error).

Answer: B.

Hope it's clear.
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There are n members in a certain department, including Michael. Two re [#permalink] New post   Updated on: 17 May 2021, 20:35
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Given: There are n members in a certain department, including Michael. Two representatives are to be selected to attend a company conference.
Asked: If there are 55 possible combinations in which Michael is not selected, what is the value of n?

If there are 55 possible combinations in which Michael is not selected, number of ways 2 representatives are to be selected to attend a company conference = (n-1)C2 = (n-1)(n-2)/2 = 55

(n-1)(n-2) = 110
n-1 = 11
n = 12

IMO B

Originally posted by Kinshook on 10 May 2021, 10:09.
Last edited by Kinshook on 17 May 2021, 20:35, edited 1 time in total.
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Re: There are n members in a certain department, including Michael. Two re [#permalink] New post   10 May 2021, 13:37
Imho the answer is B.

Number of ways of not choosing Michael would be:

(n-1)C2=55

Only 12 fits the equation!
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Re: There are n members in a certain department, including Michael. Two re [#permalink] New post   16 May 2021, 22:50
The question is best solved by using options.
we know the total ways of selection for 'n' people would be nC2, and since Michael is not to be selected we remove him from the original set and then count the ways that would lead to 55.
11c2=55( this is without Michael)
so, N(total people)=11+1(i.e with Michael) =12
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Re: There are n members in a certain department, including Michael. Two re [#permalink] New post   29 Feb 2024, 01:01
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