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