Four speakers will address a meeting where speaker Q will always speak after speaker p .Then the number of ways in whcich the prder of speakers can be prepared is -

There are 5 speakers A,B,C,D and E. The number of ways in which A will speak always before B is:

In a conference 10 speakers are present . If S_1 wants to speak before S_2 and 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

In a conference 10 speakers are present.If S_(1) wants to speak before S_(2) and S_(2) wants to speak after S_(3) ,then find 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.

In a conference 10 speakers are present. If S_1 wants to speak before S_2a n dS_2 wants to speak after S_3, then find 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.

In a conference 10 speakers are present. If S_1 wants to speak before S_2a n dS_2 wants to speak after S_3, then find 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.

In a conference 10 speakers are present. If S_1 wants to speak before S_2a n dS_2 wants to speak after S_3, then find 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.