In a conference 10 speakers are present. If $S_{1}$ wants to speak before $S_{2}$ & $S_{2}$ wants to speak after $S_{3}$ , 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

^{10} C_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{10} P_{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{10} P_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{10!}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{10!}{3}$
The easiest way in these types of question is to arrange the speakers first and then multiply with fractional probability.
Speakers can be arranged in $10!$ ways while fractional probability is $\frac{1}{3}$
So, no of ways $= \frac{10!}{3}$
Hence, the answer is $\frac{10!}{3} .$

Suggest Corrections

Similar questions

Q. In a conference 10 speakers are present . if

S_{1}

wants to speak before

S_{2}

S_{2}

wants to speak after

S_{3}

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

Q. At an election meeting 10 speakers are to address the meeting. The only protocol to be observed is that whenever they speak P.M. will speak before M.P., and M.P. will speak before M.L.A. In how many ways can the meeting be addressed ?

Q. If eight persons are to address a meeting, then the number of ways in which a specified speaker is to speak before another specified speaker is?

Q. Four speakers will address a meeting where speaker

Q

will always speak after

P

. Then, the number of ways in which the order of speakers can be prepared is

Ten persons, amongst whom are A, B and C to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is

Join BYJU'S Learning Program

In a conference 10 speakers are present. If S1 wants to speak before S2 & 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

The correct option is D 10!3The easiest way in these types of question is to arrange the speakers first and then multiply with fractional probability.Speakers can be arranged in 10! ways while fractional probability is 13So, no of ways =10!3Hence, the answer is 10!3.

In a conference 10 speakers are present. If $S_{1}$ wants to speak before $S_{2}$ & $S_{2}$ wants to speak after $S_{3}$ , 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