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11 Apr 2024, 01:30
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B
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Difficulty:
35%
(medium)
Question Stats:
76% (02:31) correct
24%
(03:26)
wrong
based on 58
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There are five empty chairs in a row. If six men and four women are waiting to be seated, what is the probability that the seats will be occupied by two men and three women?
A) 4/21
B) 5/21
C) 8/21
D) 9/21
E) 11/21
A) 4/21
B) 5/21
C) 8/21
D) 9/21
E) 11/21
jack5397
Joined: 13 Sep 2020
Last visit: 14 Nov 2024
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Location: India
Concentration: General Management, Strategy
Schools: ISB '27 IIMA '26 INSEAD '26
GMAT Focus 1: 575 Q79 V79 DI77 
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GMAT Focus 3: 635 Q82 V83 DI79 
GMAT 1: 460 Q36 V18 (Online)
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11 Apr 2024, 02:18
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Selecting 2 out of 6 men - 6C2
Selecting 3 out of 4 women - 4C3
Total possibilities- 10C5
P(E) = (6C2*4C3)/10C5 = 5/21
Posted from my mobile device
Selecting 3 out of 4 women - 4C3
Total possibilities- 10C5
P(E) = (6C2*4C3)/10C5 = 5/21
Posted from my mobile device
11 Apr 2024, 02:28
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The equation to solve this will be: [(Total ways of choosing 2 men from 6)*(Total ways of choosing 3 women from 4)]/Total ways of choosing 5 people out of 10
Total ways 5 people can be chosen out of 10:
\(\frac{10!}{5!5!}\)
\(\frac{10*9*8*7*6}{5*4*3*2}\)
\(252\)
Total ways of choosing 2 men from 6:
\(\frac{6!}{2!4!}\)
\(\frac{6*5}{2}\)
\(15\)
Total ways of choosing 3 women from 4:
\(\frac{4!}{3!1!}\)
\(4\)
Putting all the info back into initial equation:
\(\frac{15*4}{252}\)
\(\frac{5}{21}\)
ANSWER B
Total ways 5 people can be chosen out of 10:
\(\frac{10!}{5!5!}\)
\(\frac{10*9*8*7*6}{5*4*3*2}\)
\(252\)
Total ways of choosing 2 men from 6:
\(\frac{6!}{2!4!}\)
\(\frac{6*5}{2}\)
\(15\)
Total ways of choosing 3 women from 4:
\(\frac{4!}{3!1!}\)
\(4\)
Putting all the info back into initial equation:
\(\frac{15*4}{252}\)
\(\frac{5}{21}\)
ANSWER B
