7 men are 5 women are seated in a row.Find the ways they can be sated so that two women are not together.
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?
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.
Thank you!
Unfortunately they haven't given answers for this one.
Yes, you are correct!
To find the number of ways the men and women can be seated such that two women are not together, you have correctly considered two cases: starting with a man and starting with a woman.
Case 1: Starting with a man
In this case, we have 7 men and 5 women. We can first arrange the 7 men in 7! (7 factorial) ways. Then, we can arrange the 5 women in 5! ways. However, since 2 women should not be together, we need to consider the positions relative to the men. We can think of the men as fixed points and the women as objects to be arranged around them with 2 empty spaces between each pair of women. This can be done in (7! * 5!) * 2! ways.
Case 2: Starting with a woman
In this case, we have 7 men and 5 women. We can first arrange the 5 women in 5! ways. Then, we need to consider the positions for the 7 men. Since 2 women cannot be together, we can place the first woman anywhere there are 7 possible options. Then, we can place the second woman in 6 remaining options, the third woman in 5 remaining options, and so on. Therefore, the number of ways for the women followed by the men is (5! * 7 * 6 * 5 * 4) * (3 * 2 * 1).
Finally, to get the total number of ways, we sum up the two cases:
Total ways = (7! * 5!) * 2! + (5! * 7 * 6 * 5 * 4) * (3 * 2 * 1)
You have correctly taken into account the possibility of reversing the way people are seated and the additional factor of (3 * 2 * 1) for the ways the 3 remaining men can be seated in the case of starting with a woman.
Well done!