Building intuition behind combinatorics formulae I always find it to remember and apply combinatorics formulae related to combinations.
Consider below formula:
$$\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$$
Here $\binom{n-1}{r-1}$ can be interpreted as selecting $r$ objects from $n-1$ objects and then adding $n^{th}$ object as $r^{th}$ object to selection. Another way to select $r$ objects from from $n$ objects is $\binom{n-1}{r}$ that is to select $r$ objects directly from $n-1$ objects and ignore the $n^{th}$ object altogether. 
Q. Is this interpretation correct? 
This interpretation is helping me to remember this relation and recall it whenever required. However there are some folrmulae for which I am not able get any intuitive interpretation. These are:


*

*$r\binom{n}{r}=n\binom{n-1}{r-1}$

*$\frac{\binom{n}{r}}{r+1}=\frac{\binom{n+1}{r+1}}{n+1}$

*$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{n-r+1}{r}$


Can you point out the combinatorial facts that these formulae trying to communicate? Or there are just mathmatical results not mirroring any significant combinatorial fact?
 A: Counting in two ways or other combinatoric arguments will work excellently for most problems of this nature. For identities like this, try finding something easy to count and showing that both sides of the equation count it.

For your initial question regarding the identity $\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$, your justification seems to be a little unclear, specifically were
I would think about partitioning $\binom{n}{r}$ depending on whether or not it contains some specific element "a". As every subset of n either has or does not have this element, $\binom{n}{r}$ will be the sum of this partition.
If it doesn't contain this element "a", there are $\binom{n-1}{r}$ ways to form the subset (since we know "a" is not in it).  
On the other hand, if it does contain "a", there are $\binom{n-1}{r-1}$ ways to form this subset (since we already know one of the elements in it).
So, 
 $$\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$$

For (1):
Consider forming a committee of size r from a class of n students where one student of the committee is designated as the president.
On one hand, we can first form the committee ($\binom{n}{r}$ possibilities) then choose one member from this committee to be the president (r possibilities), so there are $r\binom{n}{r}$ ways to do this in total.
On the other hand, we can first choose the president from the class (n possibilities) then form the remainder of the committee ($\binom{n-1}{r-1}$ possibilities, so there are $n\binom{n-1}{r-1}$ ways to do this.  Thus,
$$r\binom{n}{r}=n\binom{n-1}{r-1}$$

For (2) and (3):
It often makes more combinatoric sense to consider expressions with nothing in the denominator, so look at $(n+1)\binom{r}{n}=(r+1)\binom{n+1}{r+1}$ instead. 
Notice that this is the same as (1), just letting $n=n+1$ and $r=r+1$, so the proof will be equivalent.
The same thing can be done for (3) and another poster gave a nice proof of the identity you asked about.
A: We'll find different ways to paint $r$ boxes out of $n$ blue, of which exactly one box is extra-blue.

Paint $r$ boxes out of $n$ blue ($\binom{n}{r}$ possibilities), and paint one of those extra-blue ($r$ possibilities). Another way to do this is picking one out of $n$ to paint extra-blue ($n$ possibilities) and then picking $r-1$ to paint normal blue ($\binom{n-1}{r-1}$ possibilities). A third way to do this, is first choose $r-1$ boxes to paint blue ($\binom{n}{r-1}$ possibilities) and then picking the one that'll be extra-blue (and that's $n-(r-1)=n-r+1$ possibilities). Thus, $$r\binom{n}{r}=n\binom{n-1}{r-1}=\binom{n}{r-1}(n-r+1)$$

Now we can derive property 1, 2 and 3 from these three expressions.
