mals24 wrote:@4meonly
Your reasoning is partially right. You haven't completed the solution.
In order to find out the number of ways John and Susan cannot sit together we can subtract the total number of arrangements possible and number of ways john and susan can sit together.
Total arrangements possible
BGBGBGBG
4*4*3*3*2*2*1*1 = 4!*4! = 576
You can also have GBGBGBGB = 4!*4! = 576
Hence total arrangements = 576 + 576 = 1152
Number of ways John and Susan can sit together
Let JS be one entity
Total arrangements JS BGBGBG = 1*3*3*2*2*1*1 = 3!*3! = 36
JS BGBGBG can be arranges in 7 ways = 36*7= 252
We can also have JS GBGBGB = 7*3!*3! = 252
Total arrangements = 252+252 = 504
2*7*3!*3! = 504 is the number of ways John and Susan can sit together.
We need to subtract this from the total number of alternative arrangements possible
Total number of arrangements possible in which John and Susan cannot sit together =1152 - 504 = 648.
Mals,
Good work. What most of the people did not see in this problem was John and Susan can sit in 7 different positions, and since they can switch = 14 ways
JS XX XX XX
XJ SX XX XX
XX JS XX XX
XX XJ SX XX
XX XX JS XX
XX XX XJ SX
XX XX XX JS
so it comes down to:
2x4!x4! - 2x7x3!x3! = 648
