# How many different arrangements are there if Bob and Sally must always be seated next to each other?

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.

• What have you tried? Where are you stuck? – Arthur Dec 7 '18 at 9:43
• 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 – user607735 Dec 7 '18 at 9:45
• Why not try to get an answer for $n = 0, 1$ or $2$ first to get a feel for what's going on? – Arthur Dec 7 '18 at 9:47
• Is the problem stated correctly, so one seat remains empty? – Christoph Dec 7 '18 at 9:53
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Dec 7 '18 at 10:07

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.

• yes so it could be writing as n+2 C n+1 – user607735 Dec 7 '18 at 9:55
• Writing what as $\binom{n+2}{n+1}$? The number of ways to seat the people is something else. – Christoph Dec 7 '18 at 9:56

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.

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 $$n!\times (n+2)\times 2\times(n+1)=2\times(n+2)!$$

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