# Combinatorics: Things always included together

Find the number of permutations of $n$ different things taken $r$ at a time so that two particular things are always included and are together?

Including two things initially, i have $(n-2)$ things from which I can choose $(r-2)$ things. Hence $\,^{(n-2)}C_{(r-2)}$ denote the combinations which can be arranged in $(r-2)!$

The two things can be interchanged in 2! ways within themselves.

Furthermore, when $r$ things are selected, I have $r+1$ ways where I can insert two things which have to remain together.

So I get $\,^{(n-2)}C_{(r-2)} \cdot (r-2)! \cdot 2! \cdot (r+1)$

Am I doing the right thing?

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You don't have $r$ things into which to put the two particular ones, you only have $r-2$ things. That gives you $r-3$ spaces in between the $r-2$ other things, plus the space at the beginning and end, for a total of $r-1$ locations where you can insert the two things that have to be together. – Arturo Magidin Jun 5 '12 at 18:59
A slightly different way. Tie together the two people who must be together. Then we have $r-1$ "people" who can be permuted in $(r-1)!$ ways. Untie them, they can permute themselves in $2!$ ways. So get $\binom{n-2}{r-2}(r-1)!2!$. – André Nicolas Jun 5 '12 at 19:15
Elegant solution Andre, thanks a lot :) – Karan Jun 6 '12 at 9:25

You now need to permute $(r-2)$ things from $(n-2)$ things as you identified, so we have $\,^{n-2}P_{r-2}$ ways of doing this. However, we have to consider two more factors, firstly, the fact that we can arrange the two items that must be next to each other in $2!=2$ ways (as you already stated), and secondly that we can put this pair of items in any of the $(r-2+1)=(r-1)$ places in the set.
$$\,^{n-2}P_{r-2} \cdot 2! \cdot (r-1)$$
N.B: $$\,^{n}P_{r}\equiv {(n)}_{r}\equiv \frac{n!}{(n-r)!}, \forall r\le n$$ And: $$\,^{n}P_{r}\equiv0, \forall r \gt n$$