As a homework question, I am asked to answer the following question:
In how may ways can $8$ people be seated in a row if
$(a)\,$ $5$ are women and they must sit next to one another?
$(b)\,$ there are $4$ married couples and each couple must sit together?
$\mathbf{(\mathit{a})}$ The way I approached the problem is by considering the 5 women seating next to each other as a single element, call it $W$, which leaves me with $3$ other elements $M_1,M_2$ and $M_3$. Therefore, by defining $k$ as the set containing the above elements, I have: $$k=\left\{W,M_1,M_2,M_3\right\}.$$ The number of ways that the elements in $k$ can be arranged is $4!$. Now, defining $W=\left\{W_1,W_2,W_3,W_4,W_5\right\}$, the women can be arranged with each other in $5!$ ways. Therefore, the solution, in my opinion, should be $5!\cdot4!=2880$. However, according to the instructor it should be $5760$.
$\mathbf{(\mathit{b})}$ With the same logic as before, the $4$ married couples can be arranged with each other in $4!$ ways and each couple can be arranged in $2!$ ways. Being that there are $4$ couples, the solution is $4!\cdot2!\cdot2!\cdot2!\cdot2!=4!\cdot2^4=384.$ Once again, my solution does not match with the one given to me, which is $192$.
It is not clear to me what I am doing wrong since I thought of these problmes as being very trivial. Any help is appreciated! Thank you!