How many different ways can 5 people--A,B,C,D and E-- sit in a row at a movie theater if 

(a) A and B must sit together; 

Then we have 3 single people and 1 pair of people to seat:

(AB), C, D, E  

That's 4 things to place in a row,  4P4 or 4! or 4*3*2*1 or 24 ways
to seat these.

But the pair could also be in reverse order. So we also have these
4 things to place in a row:

(BA), C, D, E  

That's 4 more things to place in a row,  which is also 4P4 or 4! 
or 4*3*2*1 or 24 more ways to seat these.

So the total is 4!*2 = 24*2 or 48 ways.

(b) C must sit to the right of, but not necessarily next to B; 

We can choose a pair of seats 5C2  or 10 ways, each time putting
A left of C. For each of those 10 ways, there are 3x2x1 or 3P3 or 3! 
of 6 ways to seat the other three people in the remaining 3 seats.

So that's 5C2*3! = 10*6 = 60 ways. 

(c) D and E will not sit next to to each other? 

All possible seatings are 5!, and from that we subtract the number of
ways D and E can sit together.  But that is the same as the number of ways
A and B can sit together, which we have found to be 48 ways in part (a).

So that is 5!-48 = 120-48 = 72 ways.

Edwin