Answer is 4!*5! because there are 4 women which will occupy 2,4,6,8th positions therefore 4! and 5 men will occupy 1,3,5,7,9th positions therefor 5!

Problem: It is required to seat 5 men and 4 women in a row so that women occupy the even places. How many such arrangements are possible?

Solution:
To solve this problem, we can break it down into multiple steps:

Step 1: Identify the number of even positions in a row
In this case, there are a total of 9 positions in the row. Out of these, the even positions are 2, 4, 6, and 8.

Step 2: Calculate the number of ways to arrange the women in the even positions
Since there are 4 women, we need to calculate the number of ways to arrange them in the 4 even positions. This can be done using the concept of permutations.

The number of ways to arrange 4 items in 4 positions is given by 4P4 = 4!/(4-4)! = 4! = 4 x 3 x 2 x 1 = 24.

Step 3: Calculate the number of ways to arrange the men in the remaining positions
After arranging the women in the even positions, we are left with 5 men and 5 positions (including the odd positions). We need to calculate the number of ways to arrange the men in these positions.

The number of ways to arrange 5 items in 5 positions is given by 5P5 = 5!/(5-5)! = 5! = 5 x 4 x 3 x 2 x 1 = 120.

Step 4: Calculate the total number of arrangements
To find the total number of arrangements, we need to multiply the number of ways to arrange the women (Step 2) by the number of ways to arrange the men (Step 3).

Total number of arrangements = 24 x 120 = 2880.

Therefore, there are 2880 possible arrangements for seating 5 men and 4 women in a row so that women occupy the even places.