An interesting probability problem

[billzhao]

Find the number 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.

My thinking is: I put John and Susan together as (JS) or (SJ)
there are eight categories of combinations as below:

(JS)BGBGBG (SJ)GBGBGB
BG(JS)BGBG GB(SJ)GBGB
BGBG(JS)BG GBGB(SJ)GB
BGBGBG(JS) GBGBGB(SJ)

So the number of combination is: 2*4!*4! - 8*3!*3! Is it correct?

The answer is 2*4!*4! - 14*3!*3!

Thanks.

[fleurdelisse]

You under-counted the number of times john and susan can be next to each other, it is 14. 7 for when you start the seating with a boy and 7 for when you start the seating with a girl.

you counted in pairs of (1,2), (3,4), etc. and did not account for susan and john to be in seats (2,3) for example. Do the counting again, and you'll see what I mean.

Reasoning for the whole answer, for the other readers:

Total number of ways that 4 boys an 4 girls can be seated alternatively is:
2*4!*4!, since you have 4! for seating girls and 4! for seating boys, and you multiply by 2 to account for the fact that the first seat can be either a boy or a girl

and then you remove the possibilities of having John and Susan together, which is in 14 different ways (count only for when a boy is in the first seat for example, you'll get 7 and then multiply by 2). And for each of the 14 different ways, you can seat the remaining girls and boys in 3!*3! ways

So end result: 2*4!*4! - 14*3!*3!

[fleurdelisse]

so, following your method of counting, it would be like this (in bold are those you did not account for):

(JS)BGBGBG (SJ)GBGBGB
B(SJ)GBGBG G(JS)BGBGB
BG(JS)BGBG GB(SJ)GBGB
BGB(SJ)GBG GBG(JS)BGB
etc.

[billzhao]

i got it. thanks a lot!

[Uri]

Is there any other way to find out the solution, without writing down the possible arrangements?