Question

in a conference 10 speakers are present if s1 want speak before s2 & s2 wants to speak after s3 , then the no. of ways all the 10 speakers can give the speeches with above restriction , if the remaining 7 speakers have no objection to speak at any number is?

Answer 1

This is a question on permutation and combination.

There are 7 speakers where the restrictions do not apply.

For the first three speakers we have :

S1, S3, S2

S3 can also speak before S1

There are two possibilities for S1 and S3 to take place and there is no third possibility.

For S2 it has to remain in the same position.

Now we have :

10! / (10 - 7)!

10!/3!

= 10!/3!

Answer 2

Here the following attachment may help you