[Math] Traditional combination problem with married couples buying seats to a concert.

combinationsprobability

Three married couples have bought $6$ seats in a row for a concert. How many ways can they be seated if no man sits next to his wife.

I have worked through this problem and have got the correct answer. The problem I am having is, I can't, for the life of me, wrap my head around how to find the number of ways $1$ couple sits together.

$6!$ ways to order everyone $= 720$

All couples sit together: $3!(2!)^3=48$. Three different couples can be orientated $3!$ ways and each couple can be orientated $2!$ ways. Makes perfect sense to me.

Two couples sit together: $3(3)(2!)^4=144$. Pick which couple won't sit together, $3$. There are $3$ ways they cannot sit together and $2!$ ways to order them. $2!$ ways to order the remaining two couples and $2!$ ways to order which seat man or wife takes for both of them. Okay, great, I got that.

One couple sits together: I fumbled around with this until I came up with the $288$ I needed to get the correct answer: $3(2!)^2(4!)$. But, I have absolutely no idea how to make sense of it as I did above for the other possibilities. Pick a couple, $3$, order them $2!$. Pick a second couple $2$, then next to him can't be his wife, but it can be any of the other $3$. That gives you the $3(2!)^2$ and $3$. But we still need $4$ and $2$ of the $4!$. I can't reason out how we get to that conclusion.

Final answer: $720-48-144-288=240$. There are $240$ ways for no couple to sit together.

If someone could please break it down, Barney style, how one couple sitting together must be ordered it would be fantastic. I'm having a serious mental block.

Best Answer

You could do it this way.

  • Couple sits at the end, so the arrangement must be $X_1X_2Y_1Z_1Y_2Z_2$. The number of choices for the six people, going left to right, are $6,1,4,2,1,1$, total $48$.
  • One away from the end: $Y_1X_1X_2Z_1Y_2Z_2$, choices $6,4,1,2,1,1$, total $48$.
  • Sitting in the middle: $Y_1Z_1X_1X_2(any)(any)$, choices $6,4,2,1,2,1$, total $96$.
  • The reverse of the first two cases, total $96$.

This gives your $288$.

If you wanted, you could instead do the whole thing by inclusion/exclusion. Let $A_1$ be the set of arrangements in which the first couple sit together, $A_2$ and $A_3$ likewise. Then the number we need is $$\eqalign{|\overline A_1\cap\overline A_2\cap\overline A_3| &=|{\cal U}|-|A_1\cup A_2\cup A_3|\cr &=|{\cal U}|-|A_1|-|A_2|-|A_3|+|A_1\cap A_2|+|A_1\cap A_3|\cr &\qquad\qquad\qquad\qquad\qquad\qquad{}+|A_2\cap A_3|-|A_1\cap A_2\cap A_3|\ .\cr}$$ The total number of arrangements of $6$ people is $6!\,$. To count $A_1$,

  • pick the places for the first couple to sit: as they are sitting together, this amounts to arranging the letters $PPPPC$, where $P$ is a person and $C$ is a couple. . . . . there are $5$ ways to do this;
  • pick which way round the members of the couple sit. . . . . $2$ ways;
  • order the other $4$ people. . . . . $4!$ ways.

So $|A_1|=5\times2\times4!\,$, and clearly $|A_2|,|A_3|$ are the same. For similar reasons, starting by ordering $PPC_1C_2$ we have $$|A_1\cap A_2|=P(4,2)\times2^2\times2!\ ,\quad |A_1\cap A_2\cap A_3|=3!\times2^3$$ and so our answer is $$6!-3\times5\times2\times4!+3\times P(4,2)\times2^2\times2!-3!\times2^3 =6!-6!+288-48=240\ .$$

Related Question