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10 May 2021, 08:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:03) correct
32%
(02:11)
wrong
based on 204
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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
A. 11
B. 12
C. 15
D. 18
E. 22
16 May 2021, 02:00
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Bunuel
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.
General Discussion
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
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
