At first, we'll place the women
Now, there are $10$ women
$10$ women will occupy $10$ places
& they can arrange their places in $10!$ ways
We can see that, now there are $11$ vacant places.
& there are $6$ men
So, Among $11$ places $6$ men can sit in $^{11}P_6$ ways [Choose $6$ places from $11$ places & then arrange the places]
∴ Total Ways = $10! \times ^{11}P_6$ ways
$\qquad \qquad = 10! \times \dfrac{11!}{5!}$
$\qquad \qquad = 1,20,708,40,32,000$ ways. $\rightarrow$ This is the answer.
But, you want to do this in another way -
which is to divide the women into $5$ groups, where $1$ group will contain $6$ women, & the remaining $4$ groups will contain $1$ women each.
So, let's assume $1^{st}$ group will having $6$ women
$2^{nd}$ group will contain $1$ women
$3^{rd}$ group will contain $1$ women
$4^{th}$ group will contain $1$ women
$5^{th}$ group will contain $1$ women
Now, $5$ groups can arrange in $5!$ ways
from $10$ women we've to choose $6$ women who'll be in $1^{st}$ group, that can be done in $^{10}C_6$ ways
& the group ($1^{st}$ group ) which has $6$ women, in that group $6$ women can arrange their places in $6!$ ways
The scenario will look like-
Now, vacant places are-
= $11$
∴ $6$ men can sit in $11$ places in $^{11}P_6$ ways [first, they'll choose $6$ places from $11$ places & then they arranged their position.]
∴ Total Ways = $5! \times ^{10}C_6 \times 6! \times ^{11}P_6$
$\qquad \qquad = 5! \times 210 \times 6! \times ^{11}P_6 $
$\qquad \qquad = 6.03542016e12$
But we'll not use this method to solve this problem.
It'll give us a different solution as in the question there is not mentioned anything whether women are in group or not
& when we divide 10 women into 5 groups, it will increase the total no. ways by $5$ times
[$1207084032000 \times 5 = 6.03542016e12$]