Has anyone studied this binomial like coefficients before? Consider the following identities.  
$\dfrac{n}{n-r}\dbinom{n-r}{r}=\dfrac{n}{r}\dbinom{n-r-1}{r-1}$  
$\dfrac{n-1}{n-r}\dbinom{n-r}{r-1}+\dfrac{n}{n-r}\dbinom{n-r}{r}=\dfrac{n+1}{n+1-r}\dbinom{n+1-r}{r}$       
I studied binomial like coefficient and find these two new identities. I want to know has any one studied about the binomial like coefficient $\dfrac{n}{n-r}\dbinom{n-r}{r}$ or is there any combinatorical meaning for this?
 A: All of these pretty much come from manipulating
$$
\binom{n}{r} = \frac{n!}{(r!)(n-r)!} = \frac{(n)(n-1)\dotsm(n-r+1)}{r!}.
$$
A: Here's a more general equality, and the combinatorial reason for this type of equality.  Let $n\geq k \geq r$ be non-negative integers.  Then:
$$\binom{n}{r} \binom{n-r}{k-r}=\binom{k}{r}\binom{n}{k}.$$
Proofs:
1) Consider the set $[1,n]=\{1, \cdots, n\}.$  Let 
$${\mathcal S}=\{(S,T)\subset [1,n]\times[1,n] \mbox{  s.t. } |S|=k, |T|=r, T \subset S\}.$$
One may choose first $S$ and then its subset $T$, which yields the count
$$\binom{n}{k} \binom{k}{r},$$ or alternatively choose $T$ first, and then choose $S\setminus T$, which yields:
$$\binom{n}{r} \binom{n-r}{k-r}.$$
2)  Consider the expression $\frac{1}{r!}\left(\frac{d}{dx}\right)^r(1+x)^n.$  Differentiating the Newton binomial termwise yields 
$$\sum_{k\geq r} \binom{n}{k} \binom{k}{r} x^{k-r},$$
whereas, if we differentiate  without expanding, we obtain
$$\frac{1}{r!}\left(\frac{d}{dx}\right)^r(1+x)^n=\binom{n}{r} (1+x)^{n-r},$$
 which in turn yields  (our index will run from $r$ to $n$):
$$\sum_{k\geq r} \binom{n}{r} \binom{n-r}{k-r}  x^{k-r},$$
and the identity follows.
Of course, one can simply operate the factorials, though that one is slightly less exciting :)
The genesis of the formula becomes apparent with any of the above proofs.
A: Consider $r\binom{k}{r}$ and $k\binom{k-1}{r-1}$
The first is the number of options for choosing $r$ elements from a set of size $k$, and then choosing a "leader" from that set (there are exactly $r$ options to do that)
The second is the number of options to first choose a "leader" from the set of $k$ elements, and then from the remaining $k-1$ elements, pick $r-1$ elements.
Both are the number of ways to choose a set of $r$ elements from a set of $k$ elements, and distinguish one element in that set.
So bijectively, they are the same.
For $k=n-r$, we get $r\binom{n-r}{r}=(n-r)\binom{n-r-1}{r-1}\Rightarrow \frac{n}{n-r}\binom{n-r}{r}=\frac{n}{r}\binom{n-r-1}{r-1}$
I can't think of a good direction to bijectively proof as $\frac{n}{r}$ is not necessarily an integer.
Edit:
For the second part, we will have to use some algebra
Note that $\binom{n-1}{r}+\binom{n-1}{r-1}=\binom{n}{r}$: The right side is the number of ways of picking $r$ elements from a set of size $n$, and the left is the number of ways of picking a set of $r$ elements from a set of size $n$, when the first is not picked, plus the number of ways to pick $r$ elements when the first must be picked, so they are the same.
Therefore the left side of your expression can be seen as:
$\frac{n-1}{n-r}\binom{n+1-r}{r}+\frac{1}{n-r}\binom{n-r}{r}$
We get that your claim is the same as $(n+1-r)\binom{n-r}{r}=((n+1)(n-r)-(n-1)(n+1-r))\binom{n+1-r}{r}=((n^2+n-rn-r)-(n^2-n+n-1-rn+r))\binom{n+1-r}{r}=(n-2r+1)\binom{n+1-r}{r}$
Or
$(n+1-r)(\binom{n+1-r}{r}-\binom{n-r}{r})=r\binom{n+1-r}{r}$
From the previous argument, the left side here is
$(n+1-r)\binom{n+1-r}{r-1}$
Finally, we are done, as the left side is:
picking a distinct element, and another $r-1$ elements
and the right side is:
picking $r$ elements, and picking a distinct elements from them
Therefore they are equal
A: After a very long time I found a solution for this question. I think it is worth to post as an answer. 

$$\dfrac{n}{n-r}\dbinom{n-r}{r}$$ is the number of ways selecting $r$ objects, from $n$ objects arranged in a circle so that  no two are consecutive.

Proof:
There are $n$ ways to choose the first object. We can choose remaining $r-1$ objects among $n-1-((r-1)+r)=n-2r$ objects. Furthermore, there are $\dbinom{n-r-1}{n-2r}$ ways to select those $n-2r$ positions. We must now multiply this by $n$ choices for the location of the first object and divide by $r$. Since we could pick the same group again by starting with a different member, and this can happen in $r$ ways. Hence $\dfrac{n}{r}\dbinom{n-r-1}{r-1}=\dfrac{n}{n-r}\dbinom{n-r}{r}$ ways to do this according to required conditions. 
I found this proof from this nice article.
