0
$\begingroup$

There 14 women and 8 men and they want to sit on a circular table. In how many ways can they sit so that no 2 men are sitting next to each other?

There are $(14-1)!=13!$ ways in which women can sit.

Then there are $\binom{14}{8}$ so that men can sit, or not?

So do we get the answer by multiplying these two ?

$\endgroup$
  • $\begingroup$ I think derangement will work here. You keep subtracting no. of ways in which 2,3,4.... men can sit together. $\endgroup$ – Lord KK Jun 7 '18 at 7:10
2
$\begingroup$

First of all select position for men. Men can sit in $\binom{14}{8}$ position .

Red one are position for women and green are AVALIABLE position for men Then both women and men can permute in $(14-1)!$ and $(8)!$ ways respectively.

Final answer=$$\binom{14}{8} \times 13! \times 8!$$

$\endgroup$
  • $\begingroup$ Ah ok! But this result is not at one of the choices. Is there maybe a typo? $\endgroup$ – Mary Star Jun 7 '18 at 7:06
  • 1
    $\begingroup$ may be .I think my solution is correct!! $\endgroup$ – laura Jun 7 '18 at 7:09
  • 1
    $\begingroup$ @MaryStar: i had made a mistake there ..check my answer now! $\endgroup$ – laura Jun 7 '18 at 7:19
  • $\begingroup$ I understand how you get that result! The possible result are $$ a) \frac{14!\cdot 15!}{7!} \ \ \ \ \ b) \frac{13!\cdot 14!}{6!} \ \ \ \ \ c) 22!-8! \ \ \ \ \ d) 14!\cdot 8!$$ So, your result is not equivalent of one of these, is it? $\endgroup$ – Mary Star Jun 7 '18 at 7:35
  • 1
    $\begingroup$ So, you mean that there are $\binom{14}{8}\cdot 8!\cdot 13!$ ways, right? @N.F.Taussig $\endgroup$ – Mary Star Jun 7 '18 at 9:16
1
$\begingroup$

I preassume that the persons are distinguisable and the seats are not.

Place one of the men at the table. Starting at his left hand there comes a sequence of the form:$$.m.m.m.m.m.m.m.$$where every dot stands for at least one woman and every $m$ for one man.

This comes to finding the number of sums $w_1+\cdots+w_8=14$ where the $w_i$ are positive integers, or equivalently to finding the number of sums $v_1+\cdots+v_8=6$ where the $v_i$ are nonnegative integers.

Without distinghuishing persons and applying inclusion/exclusion we find that this can be achieved on $\binom{13}7$ ways. There are $7!$ orders for men and $14!$ for women, so the final answer is:$$\binom{13}77!14!=\frac{13!14!}{6!}$$If also seats are distinguisable then this must be multiplied with factor $22$ corresponding with the number of seats available for the man that was placed at the table as first.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.