7 men are 5 women are seated in a row.Find the ways they can be sated so that two women are not together.

Question

Starting with a man,

the ways=(7!*5!)*2!,multiplied by 2! because we can reverse the way seated.

Starting with a woman,

ways=(5!*7*6*5*4)*(3*2*1),multiplied by (3*2*1) because the way the 3 men seated in the back can be changed in 3! ways.

Total is the sum of the two above.

Am I correct?

Answer 1

Your solution:
(5!*7*6*5*4)*(3*2*1)+7!*5!*2!*2!
=604800+2419200
=3024000

The first term is basically 5!7! which is the number of ways each gender can be permuted separately, and I agree with that, however they are seated.

Now coming to the restriction, which is usually solved by seating the men first in alternate seats and leaving an empty seat between adjacent men. This can be represented as
*M*M*M*M*M*M*M*
Ladies with then take turns grabbing their favourite seats. In the end, the three extra chairs are removed.

Since there are C(8,5) ways to seat the ladies, multiplied by 5!7! to account for permuting each gender, we get
Number of ways
=7!5!C(8,5)
=5040*120*56
=33868800

See if this checks with your book answer.

Answer 2

Thank you!
Unfortunately they haven't given answers for this one.