How do you arrange men and women in a line?

Think by using your hint and allowing men and women to be placed in spaces along the line. Also try to begin by placing 6 men and 6 women firstly. You will get something like the configuration:

How many ways can you seat 6 men and 6 women?

Therefore the total of ways to seat 6 men and 6 women such that no men sit together = 2* (6!) (5!) = 172,800. If there were to be no constraints regarding the seating of 6 men and 6 women the permutations of occupying 12 chairs in a circular arrangement would be 11!=39,916,800.

How many ways can men and women sit at one table?

this being the case it sets up the same problem where six men sit at one table and six women at a second table. In each case the men and women each can be seated in 6 factorial ways, 6x5x4x3x2x1, which is 720 ways for the men and 720 ways for the women.

How many men sit in positions in NG 6 men and 5?

Men sit in positions 1 3 5 7 9 11. Number of ways of seat in ng 6 Men & 5 Women = 6P6 × 5P5 = (6×5×4×3×2) × (5×4×3×2) = 720×120= 86,400 Answer. so total number of ways in which men and women can sit alternately so that no two women can sit togather

How many ways of seat in NG 6 men & 5 women?

Positions 1 to 11 from left to right. So Women sit in positions 2 4 6 8 10. Men sit in positions 1 3 5 7 9 11. Number of ways of seat in ng 6 Men & 5 Women = 6P6 × 5P5 = (6×5×4×3×2) × (5×4×3×2) = 720×120= 86,400 Answer.

How many ways can you arrange $6$ men in $11$ possible places?

A man can be inserted in any one of the eleven positions marked off with a ‘*’, and this will ensure no two men stand next to each other. Hence, we need to find how many ways we can arrange $6$ men in the $11$ possible places: this is given by $$ ^{11}P_6 = 11 imes 10 imes 9 imes 8 imes 7 imes 6 = 332640.$$

How many gaps are there among 5 women in a family?

There are 6 gaps among 5 women and barring the end gaps, all other gaps should always remain filled with atleast a man.