Question 251091: 5 people (A,B,C,D,E)sit together in a theater. How many ways can they be arranged if personB is seated to the right of, but not necessarily next to, personC???
How would you figure this out????
Answer by palanisamy(496)   (Show Source):
You can put this solution on YOUR website! 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
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