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siddharthkapoor
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29 Aug 2020, 01:52
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:28) correct
27%
(01:38)
wrong
based on 165
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A certain company will select 2 of 8 employees to fill the positions in the marketing department and 1 of 6 employees to fill the positions in the finance department. If none of the employees are suitable for the positions in both departments, how many different ways are there for 3 employees to fill 3 positions?
(A) 34
(B) 68
(C) 140
(D) 168
(E) 336
(A) 34
(B) 68
(C) 140
(D) 168
(E) 336
yashikaaggarwal
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29 Aug 2020, 04:26
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Selection of employees is mutually exclusive. No two employees are same and eligible for any other position.
The employees are to be selected from total of 8 employees and,
The three employees can be selected in 8C3 ways
=> 8*7*6*5!/5!*3*2*1
=> 56
These 3 people can be further selected into 3 ways. = 56*3 = 168
Answer is D
Posted from my mobile device
The employees are to be selected from total of 8 employees and,
The three employees can be selected in 8C3 ways
=> 8*7*6*5!/5!*3*2*1
=> 56
These 3 people can be further selected into 3 ways. = 56*3 = 168
Answer is D
Posted from my mobile device
29 Aug 2020, 19:00
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In combinations (this is a sum based on choices), we focus on what is asked for .
Please note that \(^nC_r = \frac{n!}{(n \space - \space r)! \space * \space r!)}\)
Out of 8 employees 2 have to be chosen for the marketing department = \(^8C_2 = \frac{8!}{(8 \space - \space 3)! \space * \space 2!)} = \frac{8 \space * \space 7}{2}\) = 28 ways
From the remaining 6 employees 1 person is to be selected to the finance department = [m]^6C_1 = 6 ways
Therefore total possible combinations = 28 * 6 = 168
Option D
Arun Kumar
Please note that \(^nC_r = \frac{n!}{(n \space - \space r)! \space * \space r!)}\)
Out of 8 employees 2 have to be chosen for the marketing department = \(^8C_2 = \frac{8!}{(8 \space - \space 3)! \space * \space 2!)} = \frac{8 \space * \space 7}{2}\) = 28 ways
From the remaining 6 employees 1 person is to be selected to the finance department = [m]^6C_1 = 6 ways
Therefore total possible combinations = 28 * 6 = 168
Option D
Arun Kumar
