Arrangements in a circle

Question

There are $10$ people sitting in a circle. $4$ men and $6$ women.How many arrangements will be there if men do not appear consecutively?

Answer 1

Letting the $6$ women sit down will be $\frac{6!}{6} = 120$ different ways.

Then just insert the $4$ men into the $6$ gaps, so that no two men are in the same gap. This results in a factor of $6 \times 5 \times 4\times 3 = 360$.

So the total is $120 \times 360 = 43200$.

Answer 2

HINT:

"Glue" a man to the right of each woman consider these pairs as one unit. This way you ensured that when you arrange these units and the two remaining woman there wont be consecutive men. Use circular permutation.

Hope this helped

Answer 3

Men: Andrew, Bart, Charles and Damocles, Destroyer of Worlds

Women: Alice, Barbara, Cynthia, Diana, Elsa and Francine.

Put Andrew somewhere in the circle. There's seven slots left for the remaining three men (as you can't put them in Andrew's spot or next to him). There are ten different ways to arrange the three remaining men:

M X M X M X X

M X M X X M X

M X M X X X M

M X X M X M X

M X X M X X M

X M X M X M X

X M X M X X M

X M X X M X M

X X M X M X M

(look up 'stars-and-bars combinatorics' if you want to research the more general method to solve this)

We now have $3! = 6$ ways of seating the men, and $6! = 720$ men of seating the women, leading to a result of $4320 \cdot 10 = 43200$.

(Fixing Andrew's seat got rid of the permutations possible by rotating the circle)