Let n be the number of ways in which 5 men and 6 women can stand in a queue such that all the women stand consecutively. Let m be the number of ways in which the same 11 persons can stand in a queue such that exactly 5 women stand consecutively. The value of $$\frac{m}{n}$$ is

For n, we consider B as a group that has all 6 women. Total arrangement of B and 5 men is 6!. Further within B women can rearrange themselves in 6! ways.

Thus n = 6! $$\ \times\ $$6!

For m, consider a group $$B_1$$ which has 5 women and $$B_2$$ has remaining 1 woman. Arrangement of $$B_1$$,$$B_2$$ and 5 men can be done in 7! ways. From this we need to subtract the permutation in which $$B_1$$ and $$B_2$$ are together. Which is $$2!\times\ 6!$$ . Total ways $$=\ 7!\ -\ 2!\times 6!$$ .Which equals $$5\times\ 6!$$. Futhermore in $$B_1$$ the selection and arrangement of 5 women can be done in $$_5^6C\times\ \ 5!$$ = 6! ways.

m = $$5\times\ 6!\times\ 6!$$

$$\ \frac{\ m}{n}=5$$