Number of ways to arrange N things in R numbered row such that 2 special things are selected and are not in first or last of the row

Question

From $n$ amount of things, if we take $r$ things in which $2$ special things have to be selected, how many ways can we arrange the things so that the two special things are not in the first or last of the rows?
I want to solve this problem by finding out the (total ways - ways to have the special things at first or last).

I found total ways to be

$$\frac{(n-2)!(2nr-r^2-r)}{(n-r)!}$$

But the problem becomes an algebraic mess from there. I would be forever grateful if anyone provides a solution using the total ways - first or last formula.

Edit : I have found the total amount of ways to be $$\frac{r(r-1)(n-2)!}{(n-r)!}$$ This is because the $1$st special thing can be sorted into $r$ positions , the $2$nd special thing can be sorted into $(r-1)$ positions and the other $(n-2)$ things can be sorted into $(r-2)$ positions.

Edit(2) : Guys I think I got it. The total ways can be calculated by $$\frac{r(r-1)(n-2)!}{(n-r)!}$$

Now we calculate the number of ways the special letters are at $1$st or last of the row (we name this process point two)

Now, we consider the following cases

Case : $1$

If the $1$st special letter is at the 1st of the row, the $2$nd special letter has $(r-1)$ places to be so in this case, the $(n-2)$ other things/letters have $(r-2)$ places to be in

So in this case , Number of ways = $$\frac{(r-1)(n-2)!}{(n-r)!}$$

Case: $2$

If the 1st special letter is at the last of the row it follow the same process as case 1 . So the number of ways = $$\frac{(r-1)(n-2)!}{(n-r)!}$$

Case: $3$

In this case , we consider the $2$nd special letter being in the first of the row . The process clearly remains the same as Case 1

So, Number of ways = $$\frac{(r-1)(n-2)!}{(n-r)!}$$

Case: $4$

In this case we consider $2$nd special letter being in the last of the row . The process clearly remains the same as case 1

So, Number of ways = $$\frac{(r-1)(n-2)!}{(n-r)!}$$

Total number of ways(In these 4 cases) = Case 1 + Case 2 + Case 3 + Case 4 =$$\frac{4(r-1)(n-2)!}{(n-r)!}$$

Some of these cases overlap since when suppose $1$st special letter is at the first of the row , the $2$nd special letter can be at the last of the row . So in this scenario , Case 1 and Case 4 overlap . In order to get rid of this overlapping we consider the following cases

Case $5$

This is the case of the $1$st special letter being in the $1$st of the row and the $2$nd special letter being at the last of the row. Since these two are fixed we have $(n-2)$ other things which can go in $(r-2)$ places. So number of ways = $$\frac{(n-2)!}{(n-r)!}$$

Case $6$

This is the case of the $2$nd special letter being in the $1$st of the row and the $1$st special letter being in the last of the row . This follows the process of case 5 . So number of ways = $$\frac{(n-2)!}{(n-r)!}$$

Adding these two cases we get the total number of ways the cases can overlap = $$\frac{2(n-2)!}{(n-r)!}$$

We subtract these two cases from the first 4 cases =

$$\frac{(4r-6)(n-2)!}{(n-r)!}$$

This is Point $2$ , So we subtract this from the total number of ways and get the result of =

$$\frac{(r-2)(r-3)(n-2)!}{(n-r)!}$$

This is what the question asked

Answer 1

The posting's use of the fraction $~\displaystyle \frac{(n-2)!}{(n-r)!}~$ suggests that all $n$ items are to be considered distinct from each other, and that (therefore), the order that the items appear in the row is pertinent.

I am assuming that there are exactly $r$ positions in the row, where the $r$ items are to be placed.

My approach is to take the distribution in stages, using Inclusion-Exclusion to polish the answer. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

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First, you have to select $(r-2)$ non-special items to accompany the $2$ special items.

This can be done in

$$\binom{n-2}{r-2} ~~\text{ways}. \tag1 $$

Assume that $(r-2)$ non-special items have been selected.

For any set $E$ with a finite number of elements, let $|E|$ denote the number of elements in the set $E$.

Let $S$ denote the set of all distributions of the $[(r-2) + 2]$ selected items, where the restriction on where the two special items are placed is ignored.

Let $S_1$ denote the subset of $S$, where the 1st special item occurs in either the first or last position.

Let $S_2$ denote the subset of $S$, where the 2nd special item occurs in either the first or last position.

Then, the desired computation, omitting the factor in (1) above, will be

$$|S| - |S_1 \cup S_2|.$$

By inclusion-Exclusion, this equals
$|S| - |S_1| - |S_2| + |S_1 \cap S_2|.$

Then:

• $|S| = r!.$

• $|S_1| = 2 \times (r-1)!$.
  That is, the 1st special item can go in either the first position, or the last position. Once the 1st special item is set, there are $(r-1)!$ ways of positioning the other $(r-1)$ items, which includes the 2nd special item.

• $|S_2| = 2 \times (r-1)!$.
  This follows by the same analysis as in the previous bullet point.

• $|S_1 \cap S_2| = 2! \times (r-2)!.$
  Here, it is being assumed that both the 1st special item and the 2nd special item are in violation. There are $(2!)$ ways of arranging these two special items in the first/last places on the row. Then, there are $(r-2)!$ ways of placing the remaining $(r-2)$ items.

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Putting all of this together with the initial factor in (1) above gives a final computation of

$$\binom{n-2}{r-2} \times \left\{ ~[r!] - [4 \times (r-1)!] + [2 \times (r-2)!] ~\right\} $$

$$= \binom{n-2}{r-2} \times [(r-2)!] \times \left\{ ~[r(r-1)] - [4 \times (r-1)] + [2] ~\right\} $$

$$= \frac{(n-2)!}{(n-r)!} \times [r^2 - 5r + 6].$$

This answer agrees with the final calculation in the posted question.