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There are $3$ men and $3$ women.They should be arranged in such a way that no man stands next to another man how many ways is it possible ?

I got as far as - there can either be a man standing first or a woman. so it would be either M1 W1 M2 W2 M3 W3 or W1 M1 W2 M2 W3 M3 for the first possibility,men can be arranged in 3! ways and women can also be arranged in $3!$ ways so they can be arranged in $3!*3!$ ways which is $36$ ways.

This is also the case for the second possibility so there are $36+36=72$ ways of arranging them but the answer is $144$ ways $(72\times2)$ how???

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    $\begingroup$ The condition forbids two men from standing next to each other, but two women are allowed to do so. Hence the possible gender arrangements are $MWMWMW$, $MWMWWM$, $MWWMWM$, and $WMWMWM$. Each of these arrangements extends to $3! \cdot 3! = 36$ permutations, as you explained, giving a total of $4 \cdot 36 = 144$. $\endgroup$ – Shagnik Dec 1 '16 at 7:01
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    $\begingroup$ Why post a complete and correct answer as a comment and not as an answer? $\endgroup$ – G Tony Jacobs Dec 1 '16 at 7:22
  • $\begingroup$ It looks like you misunderstood the problem. "No man next to another man" is not the same thing as "men and women alternating". For instance, consider $MWWMWM$. $\endgroup$ – Federico Poloni Dec 1 '16 at 15:23
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The situation can be described with a little help from a diagrammatic representation...

Suppose !'s are men and *'s are women. Now the following arrangements are possible:

  1. ! * ! * ! *

  2. * ! * ! * !

  3. ! * ! * * !

  4. ! * * ! * !

Now this question becomes much easier. The first arrangement is possible in $3!\times3!=36$ ways. Similarly the second arrangement is possible in $3!\times3!=36$ ways. The third one is possible in ${3 \choose 2}\times2!\times3!=36$ and similarly the fourth situation is also possible in ${3 \choose 2}\times2!\times3!=36$ ways. So the total possible ways are $4\times36=144$.

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You should be more careful re headings. Your heading contradicts the actual question !

As for the question, the men can be inserted at any $3$ of the uparrows, $\uparrow W \uparrow W \uparrow W \uparrow$

thus, using permutation notation, # of ways = $^4P_3\times{^3P_3} = 144$

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    $\begingroup$ I like this because it extends naturally to: there are n men and n women; then the result is $^{n+1}P_n\times{^{n}P_n}$. $\endgroup$ – Chas Brown Dec 1 '16 at 10:08
  • $\begingroup$ @true blue anil you got my upvote as your answer is far bettar than that of mine. $\endgroup$ – Vidyanshu Mishra Dec 1 '16 at 15:17
  • $\begingroup$ @ChasBrown: Glad you liked it ! $\endgroup$ – true blue anil Dec 2 '16 at 3:37
  • $\begingroup$ @THELONEWOLF: Thanks for the compliment ! $\endgroup$ – true blue anil Dec 2 '16 at 3:38

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