IIT JEE , AIEEE Forum - Mathematics , Algebra

ashish_banga

7 different lecturers are to deliver lectures in 7 periods of a class on a particular day. A , B and C are 3 of the lecturers. the number of ways in which a routine for the day can be made such that A delivers his lecture before B ,and B before C is

eshaan

is the answer 936

mukundmadhav

7C3 X 4!

First you choose 3 periods for the lecturers to deliver the lecture.. These three will be filled in the order A before B before C. So now permutations of this. Now you have 4 remaining periods.. These can be shared in any order by the 4 other teachers. Number of ways is 4!..
So Answer is 7!/3! =840

allamraju

Replace A,B,C by three X's.Now,Arrange these three X's and four other lecturers in an order.This can be done in 7!/3!.Now,Replace first X by A,second by B and third by C.Hence the ans is 840.

ashish_banga

allam how u considered all same i.e. X

mukundmadhav

He's done the same thing as I have.. Just denoted it differently. Don't worry about it.

allamraju

Bcoz the order of A,B,C can't be changed.If they had been diff.,then they would have been permuted too.So,we first replace all of them by a single thing and then,after permuting we replace first X by A,second by B and third by C.Hope u got it.

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Replace A,B,C by three X's.Now,Arrange these three X's and four other lecturers in an order.This can be done in 7!/3!.Now,Replace first X by A,second by B and third by C.Hence the ans is 840.

Bcoz the order of A,B,C can't be changed.If they had been diff.,then they would have been permuted too.So,we first replace all of them by a single thing and then,after permuting we replace first X by A,second by B and third by C.Hope u got it.