How many ways are there to pick r objects from n objects when repetitions are allowed and either both the first and the second objects appear exactly once or both do not appear?
So here is what I tried and I'm not sure if it's correct.
If we have n bins and each bin is an object, and we choose each object once we will have r-n, but since the first and second object appear once or not at all we have r-(n-2)+-2, at least I think. So using
$n+k-1 \choose k$
I think I can get
$n+(r-(n-2)\pm2)-1 \choose k$
Is this correct? I dont think it is, but I'm not sure how else to go about this.
We can simply separate the $2$ cases:
1 and 2 element don't appear
This means that we must choose $r$ elements with repetition, from $n-2$ elements:
$${{n-2+r-1}\choose{r-1}}={{n+r-3}\choose{r-1}}$$
1 and 2 element appear one time
This time we have to select $r-2$ elements with repetition, from $n-2$ elements:
$${{n-2+r-2-1}\choose{r-2-1}}={{n+r-5}\choose{r-3}}$$
Since the two cases are disjoint, the result is the sum of them:
$${{n+r-5}\choose{r-3}}+{{n+r-3}\choose{r-1}}$$