Problem Solving

oa coming when a few people have answered. this question from difficult math problems doc:

Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

1) in a row
2) 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
3) around a table

i)Arrange the 4boys in a row 4!
Then 4 girls can be arranged in 5 empty spaces in 5P4 ways
So total = 4! x 5P4

iii)Arrange 4 boys around a table in 3!ways
Arrange the 4girls in 4 places in 4! ways
So total = 3! x 4!

Sorry but I am not sure of the answer for (ii)

A:
1) 4! * 4! * 2
2) 4! * 4! * 2 - number of ways with John and Susan sitting together
= (4! * 4! * 2) - (7 * 3! * 3! *2)

The way that JS arrangements are found is by
bracketing J and S and considering it to be a single entity. So a
possible arrangement is (B=boy, G=girl)

(JS) B G B G B G number of arrangements is 7 x 3! x 3! = 252
(SJ) G B G B G B number of arrangements is 7 x 3! x 3! = 252


3) Fix one boy and arrange the other 3 boys in 3! ways. Arrange the 4
girls in 4! ways in the gaps between the boys.

Total arrangements = 3! x 4!

= 6 x 24

= 144

Are you sure the answer is correct for the 2nd one? I'm not sure how you got 7 again.
You shouldnt be re-arranging the boys and girls among themselves since the question says they have to be seated alternatively

Neo2000 wrote:Are you sure the answer is correct for the 2nd one? I'm not sure how you got 7 again.
You shouldnt be re-arranging the boys and girls among themselves since the question says they have to be seated alternatively

not sure about that 2nd one, this is just the OA i have. thanks for following up

thanks mark understood every word of it beautiful explanation