there is this question of permutation given to me, though i fail to understand, and only the result is given to us, there is no any explanation regarding the solution, so could you please explain to me how it came the result to be such?
Question: How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other?
Result: $8! \cdot P_{9,5}$
Thank You. With Respect Umer Selmani
On
Firstly form a line with the $8$ men. We can do this in $8!$ different ways. Then we need to place the $5$ women in between the $8$ men such that no two women are between the same two men. There are $9$ possible positions we can place the women (to the left of any man or to the right of the $8$th man). Hence we can place the $5$ women in between the men in $5!\binom{9}{5}=9P5$ ways. Combing these values we get a total number of permutations as $$8!\cdot5!\binom{9}{5}=8!\cdot9P5$$
There are $8!$ ways to arrange $8$ men in a line and since no two women can stand next to each other, there are $9$ possibilities for the $5$ women, for a total of $9P5 = P(9,5)$ possibilities as order matters. We can think of this as a stars and bars problem. If the men are represented by the symbol "#," then we have that the line can be of the form "_ # _ # _# _# _# _# _# _." In this problem, we are essentially choosing $5$ of the $9$ blanks, but since order matters, we have to multiply by $5!,$ the number of ways to order the women in each arrangement, and so we get $9P5$ possibilities.