Question:
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.
(A) 648
(B) 2/9
(C) 1/55
(D) 864
(E) 1152
SPOILER: OA:D
Solution:
First I found how many total ways there are to arrange the girls and boys.
GB GB GB GB
4B 3B 2B 1B
G4 G3 G2 G1
Four girls can sit in the first seat. If one girl sits down, 3 girls can sit in the second available seat. so forth . . .
There are 4 x 3 x 2 x 1 ways to arrange the girls
There are 4 x 3 x 2 x 1 ways to arrange the boys
We can swap the orders of each girl-boy pair like this:
BG BG BG BG
So we multiply by 2
So the total is
4!4!2
Now we subtract out the cases where John and Susan sit together.
Think of JS and a single unit.
JS BG BG BG
If JS is in the first slot, we can arrange the rest of the girls and boys in
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
We can not swap them otherwise the alternate order is broken
Now we move JS to the second slot.
BG JS BG BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
Now we move JS to the third slot
BG BG JS BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
Now we move JS to the fourth slot
BG BG BG JS
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
So in total we have:
4 x 3!3! = 144 ways
We can also arrange the group with a girl first:
SJ GB GB GB
So in total we have:
4 x 3!3! = 144 ways
So the total ways S and J can sit together is:
3!3!4 + 3!3!4 = 288 ways
The answer would be:
4!4!2 - 3!3!8 = 864
The correct answer is D.
(46)
(28)
