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16 Apr 2019, 21:47
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
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Question Stats:
75% (01:45) correct
25%
(02:11)
wrong
based on 102
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In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the beginning and at the end?
(A) 3!/5! * 7!
(B) 5!/6! * 7!
(C) 5!/3! * 7!
(D) 3!/5! * 7!
(E) 5!/7! * 7!
(A) 3!/5! * 7!
(B) 5!/6! * 7!
(C) 5!/3! * 7!
(D) 3!/5! * 7!
(E) 5!/7! * 7!
eswarchethu135
GMAT 2: 640 Q49 V27
Posts: 277
16 Apr 2019, 22:13
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B _ _ _ _ _ _ _ B
The two boys can be selected in \(^5C_2\) ways and can be arranged in 2! ways. The remaining 7 people can be selected in 1 way and can be arranged in 7! ways.
\(\frac{5!}{2!*3!}*2! * 7!\)
\(\frac{5!}{3!}*7!\)
OPTION: C
The two boys can be selected in \(^5C_2\) ways and can be arranged in 2! ways. The remaining 7 people can be selected in 1 way and can be arranged in 7! ways.
\(\frac{5!}{2!*3!}*2! * 7!\)
\(\frac{5!}{3!}*7!\)
OPTION: C
General Discussion
16 Apr 2019, 23:47
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Bunuel
In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the beginning and at the end?
(A) 3!/5! * 7!
(B) 5!/6! * 7!
(C) 5!/3! * 7!
(D) 3!/5! * 7!
(E) 5!/7! * 7!
(A) 3!/5! * 7!
(B) 5!/6! * 7!
(C) 5!/3! * 7!
(D) 3!/5! * 7!
(E) 5!/7! * 7!
total arragements of boys
5c2 *2 ways and total places left after 2 boys occupy first and last position 7
so 5c2*2*7!
IMO C 5!/3! * 7!
