Probability of $13$ men and $2$ women divided in $3$ equal groups such that no women are in same group

$13$ men and $2$ women are to be divided into $3$ groups of $5$ each. Find the probability that the women will be in different groups.
My attempt: Total number of members are $15$. Let $A,~B,~C$ be the groups. Now the total number of ways we can placed the members in the groups with required condition is $$\frac{{15\choose 5} {10 \choose 5} {5\choose 5}}{3!}=126126.$$ Now let us put all the women in a single group, without loss of generality let it be first group $A$. Then number of ways we can placed the remaining members (all men) in the groups with required condition is $$\frac{{13\choose 3} {10 \choose 5} {5\choose 5}}{2!}=36036.$$ Now the required probability that, all the women will be placed in different groups is given by $$1-\frac{36036}{126126} =\frac{5}7$$ But the answer is given as $\dfrac{2}7$ and I am not able to find my mistake. Is my procedure is wrong or I have done any mistake in my solution? Please help me to solve this question. Thank you.

Solution 1:

Foregoing binomial coefficients, assign Ann to any of $15$ available committee slots, then assign Barbara to any of the remaining $14$ slots. Of these, only $4$ will be on the same committee as Ann, so the probability the two women will be on different committees (aka groups) is $10/14=5/7$.

Solution 2:

Yes your work is correct and the answer given in the book is incorrect.

You used complimentary method. Alternatively, we can count number of ways to make three groups with women in different groups as -

$ \displaystyle {13 \choose 5} {8 \choose 4}$

Dividing by unrestricted number of ways to make three groups $ ~ \displaystyle \frac{1}{3!} {15 \choose 10}{10 \choose 5} ~ $ gives the desired probability of $ \displaystyle \frac 57$.