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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!

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    $\begingroup$ Your work seems to be correct. Notice that your instructor’s answer in $(a)$ is twice as much as yours, and in $(b)$ is half of yours. I think your professor made some mistakes! $\endgroup$ – DMH16 Jan 27 '18 at 23:35
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    $\begingroup$ $\mathbf{(\mathit{a})}$ Your answer should be the correct one. $\mathbf{(\mathit{b})}$ Your answer should be the correct one. $\endgroup$ – ArsenBerk Jan 27 '18 at 23:36
  • $\begingroup$ @DMH16 I did not notice that. Thanks ! $\endgroup$ – user372003 Jan 27 '18 at 23:43

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