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19 May 2024, 00:29
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
92% (00:49) correct
7% (01:51) wrong based on 14 sessions
History
Date
Time
Result
Not Attempted Yet
A manager is forming a 6-person team to work on a certain project. From the 11 candidates available for the team, the manager has already chosen 3 to be on the team. In selecting the other 3 team members, how many different combinations of 3 of the remaining candidates does the manager have to choose from?
A. 6
B. 24
C. 56
D. 120
E. 462
A. 6
B. 24
C. 56
D. 120
E. 462
19 May 2024, 13:17
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We have about 11 people from which a 6 man team is to be selected. Now three members are already selected from 11, so there are 3 places left on the team.
So we need to select 3 members from 8.
Number of ways we can form the team = \(\frac{8!}{(3! * 5!)}\) = 56.
Hence the answer is 56.
So we need to select 3 members from 8.
Number of ways we can form the team = \(\frac{8!}{(3! * 5!)}\) = 56.
Hence the answer is 56.
22 Jun 2024, 03:27
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Official Explanation
To determine the number of different combinations of 3 of the remaining candidates that the manager has to choose from, you first have to know the number of remaining candidates. Since you know that the manager has already chosen 3 of the 11 candidates to be on the team, it is easy to see that there are 8 remaining candidates. Now you need to count how many different combinations of 3 objects can be chosen from a group of 8 objects.
If you remember the combinations formula, you know that the number of combinations is \(\frac{8!}{3!(8-3)!}\). You can then calculate the number of different combinations of 3 of the remaining candidates as follows.
\(\frac{8!}{3!(8-3)!} =\frac{8*5*6*5!}{(3!)(5!)} = \frac{8*7*6}{6} = 56\)
The correct answer is Choice C.
To determine the number of different combinations of 3 of the remaining candidates that the manager has to choose from, you first have to know the number of remaining candidates. Since you know that the manager has already chosen 3 of the 11 candidates to be on the team, it is easy to see that there are 8 remaining candidates. Now you need to count how many different combinations of 3 objects can be chosen from a group of 8 objects.
If you remember the combinations formula, you know that the number of combinations is \(\frac{8!}{3!(8-3)!}\). You can then calculate the number of different combinations of 3 of the remaining candidates as follows.
\(\frac{8!}{3!(8-3)!} =\frac{8*5*6*5!}{(3!)(5!)} = \frac{8*7*6}{6} = 56\)
The correct answer is Choice C.
