In a restaurant, 16 men & 10 women are seated on 26 chairs at a round table. How many possible ways men are always seated together? In a restaurant, 16 men and 10 women are seated on 26 chairs at a round table. Find the total number of possible ways such that 16 men are always sitting next to each other.
I've arrived at the solution of $16!.10!$ But I suppose the answer cannot be as simple as this, is my approach and answer correct?
It is challenging for me to frame the solution correctly.

On a circular table, $n$ objects can be arranged in $(n-1)!$ ways.
First arrange all the women on the circular table.

$10$ women can be arranged in $(10-1)! = 9!$ ways on a circular table.

Now we need to arrange $16$ men such that all of them are together.
We can arrange these men in any of the $10$ gaps between the women. (Assume that we have tied up all the men together and are considering them as a single element).

For arranging this element (all the men) in $10$ gaps, there are $10$ ways.

But these men can also change their order. So, multiply $10$ with the factorial of number of men i.e. $16!$.
So the required number of ways are,
$$9! \cdot 10 \cdot 16! = 10! \cdot 16!$$
Hence your answer is correct!
