7 diff lecturers r to deliver lectures in 7 periods of a classon a particular day . A,B,C r 3 of the lecturers . d no. of ways in which a routine for d day can b made such tat A delivers his lecture bfr B and B bfr C is???

Actually its a simple multinomial problem if you can see it.

LEt the number of people before A be x1, between A and B be x2, between B and C be x3 and After C be x4.

We know the bounds on each xi is that they are non negative integers and that their sum is 4. Thus we can set up a multinomial coefficient and then permute the people!

The ans is 840 ways.If correct,I will explain.

It is similar to arranging 7 diff. things in a row such that three particular things A,B,C are in a particular order.

Let us replace A,B,C by X and arrange them,The no. of ways are 7!/3!=840 ways.

Now,replace first X by A,second by B,third by C to get the desired order.

The lectures of A,B,C can be in the same order cannot even reordered amongst themselves.Therefore it is a simple arrangement of 5 elements amongst themselves which is 5!.I think Allamraju has considered internal arrangement.As per the question the order amongst themselves should not be changed.

Correct me if wrong

No, he's not considered internal arrangement.

He's considered the trio of ABC to be in all the different places.

If the question was that ABC lecture right in the beginning(or at any fixed time), then it'll be 4! as someone said.

Does that make it clear?