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11 Feb 2021, 04:00
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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:
85%
(hard)
Question Stats:
48% (02:06) correct
52%
(02:10)
wrong
based on 119
sessions
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In a conference 10 speakers are present. If S1 wants to speak before S2 and S2 wants to speak after S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining seven speakers have no objection to speak at any number is
(A) 10C3
(B) 10P8
(C) 10P3
(D) 10!/3
(E) 10!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 10C3
(B) 10P8
(C) 10P3
(D) 10!/3
(E) 10!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
11 Feb 2021, 05:19
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Bunuel
The 10 speakers can speak in 10! ways.
However, S1 before S2 and S2 after S3 means there can be two ways S1, S3, S2 OR S3, S1, S2. But they can be arranged within themselves in 3!, out of which only these 2 are correct.
Thus the total has to be divided by 3! and multiplied by 2 => \(\frac{10!*2}{3!}=\frac{10!}{3}\)
OR
Pick 3 seats out of 10 and place S1, S2 and S3 in two ways due to the restrictions => 2*10C3=\(2*\frac{10!}{7!3!}\)
However, the remaining 7 speakers can get their turns to speak in 7! ways
Total = \(2*\frac{10!}{7!3!}*7!=\frac{10!}{3}\)
D
