<PRE><B>Question 251091</B>
5 people (A,B,C,D,E)sit together in a theater. 
Total number of arrangements = 5P5 = 5! 
First let us consider the case in which C occupies the left extreme seat.
The other four people can be arranged in 4P4 = 4!= 1x2x3x4 = 24 ways.

 Next, let C occupy the second seat.
Excluding B, the first seat can be filled by either one of A,D,E in 3 ways and the other 3 seats can be filled in 3! ways.
So, total number of ways = 3x3! = 3x3x2x1 = 18

 Next,let C occupy the third seat.
Excluding B, the first two seats can be filled by either one of A,D,E in 
3P2 = 6 ways and the other 2 seats can be filled in 2! ways.
So, total number of ways = 6x2! = 6x2x1 = 12

 Next,let C occupy the fourth seat.
Excluding B, the first three seats can be filled by either one of A,D,E in 
3P3 = 3! ways and the other last seat can be filled by B in 1 way 
So, total number of ways = 3!x1 =3x2x1x1 = 6

So the total number of ways in which B sits right of C is = 24+18+12+6 =60 

ANOTHER METHOD
5 people (A,B,C,D,E)sit together in a theater. 
Total number of arrangements = 5P5 = 5! 
There are only two possible ways.
Either B can sit to the right of C or B can sit to the left of C.
The number of ways are equal in both cases.
Therefore the total number of ways in which B sits right of C is = 5! / 2
                                                                                                                             =1*2*3*4*5/2  =60                          </PRE>