What is the solution to problem 6 in the Combinatorics seating problem?

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Hello,
so this is what I am stuck on:
In how many different ways can you seat 11 men and 8 women in a row so that no two women are to sit next to each other.

I know it's going to be combination and not permutation and that total number of seating them is 11!*8! but that is as far as I could get :frown:
 
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Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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