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How many different arrangements are there in which Bob, Sally and $n$ other people sit down in a row of $n+3$ chairs if Bob and Sally must always be seated next to each other?

I tried putting Bob and Sally next to each other in the first two chairs so then there are $2 \times n!$ arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this.

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    $\begingroup$ What have you tried? Where are you stuck? $\endgroup$ – Arthur Dec 7 '18 at 9:43
  • $\begingroup$ I tired putting bob and sally next to each other in the first two chairs so then there is 2x(n!) arrangements but then I need to move them to the 2nd and 3rd chair and so on. Not exactly sure how to do this $\endgroup$ – user607735 Dec 7 '18 at 9:45
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    $\begingroup$ Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on? $\endgroup$ – Arthur Dec 7 '18 at 9:47
  • $\begingroup$ Is the problem stated correctly, so one seat remains empty? $\endgroup$ – Christoph Dec 7 '18 at 9:53
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Dec 7 '18 at 10:07
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Hint: Consider Bob and Sally one unit taking two seats. So you have to place $n+1$ units at $n+2$ spots. In the end multiply by $2$ to account for Bob sitting on the left or right of Sally.

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  • $\begingroup$ yes so it could be writing as n+2 C n+1 $\endgroup$ – user607735 Dec 7 '18 at 9:55
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    $\begingroup$ Writing what as $\binom{n+2}{n+1}$? The number of ways to seat the people is something else. $\endgroup$ – Christoph Dec 7 '18 at 9:56
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Treat Bob and Sally as the same person (say, V) that takes up $2$ spaces, i.e. you put it in chair $m$ and then you can't put anyone in chair $m+1$. This means V cannot be in chair $n+3$. So, we place $V$ first in $n+2$ seats, and since it takes up $2$ seats, we now have $n+1$ seats for $n$ people. This means that the number of permutations is $(n+2)!$. Now, we can have $2$ "states" for $V$: one where Bob sits on chair $m$ and Sally on chair $m+1$, or Sally sits on chair $m$ and Bob sits on $m+1$. So, we multiply by $2$ to account, for a grand total of $2(n+2)!$ seatings arrangements.

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You can evaluate the arrangements separately.

So for instance, the different arrangements in which Bob and Sally sit together are $$(n+2)\times 2$$ Note that deciding whether Bob or Sally sits to the right doubles the possibilities.

Now, all possible arrangements in which $n$ different people sit (it doesn't matter whether together or not) is simply $$n!$$

All possibilities are hence $$n!\times (n+2)\times2$$ Note, however, that you also have to consider the gap; the total amount of possibilities is thus $$n!\times (n+2)\times 2\times(n+1)=2\times(n+2)!$$

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  • $\begingroup$ You could also just exchange the gap by an additional person (since there is only one) and obtain $(n+1)!$ instead of $n!$ immediately. $\endgroup$ – Christoph Dec 7 '18 at 10:00
  • $\begingroup$ @Christoph you're definitely right! $\endgroup$ – Dr. Mathva Dec 7 '18 at 10:26

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