In a Conference 10 Speakers are Present GMAT Problem Solving

Question: 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

1. \(^{10}C_3\)
2. \(^{10}P_8\)
3. \(^{10}P_3\)
4. \(\frac{10!}{3}\)
5. 10!

Answer:
Solution with Explanation:
Approach Solution (1):

Pick 3 seats out of 10 and place S1, S2, and S3 in two ways due to the restrictions

\(2*^{10}C_3=2*\frac{10!}{7!*3!}\)

However, the remaining 7 speakers can get their turns to speak in 7! Ways

Total = \(2*\frac{10!}{7!*3!}*7!=\frac{10!}{3}\)

Correct Option: D

Approach Solution (2):

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! Ways, out of which only these 2 are correct.

Thus the total has to be divided by 3! And multiplied by 2 = \(\frac{10!*2}{3!}=\frac{10!}{3}\)

Correct Option: D

“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”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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