How many combinations of $3$ natural numbers are there that add up to $30$? How many combinations of $3$ natural numbers are there that add up to $30$?
The answer is $75$ but I need the approach. 
Although I know that we can use $_{(n-1)}C_{(r-1)}$ i.e. $_{29}C_2 = 406$ but this is when $a, b, c$ are distinguishable which is not the case here.
Please explain.
EDIT
three such examples:  
$\begin{align}
1+1+28 \\
10+10+10 \\
1+2+27 \\
...
\end{align}$
 A: The ideas of this method is known as burnsides lemma in group theory. 
As you pointed out, the number of positive integer solutions to $a+b+c=30$ is ${29 \choose 2}=406$ by the stars and bars. However it over counts the total because the numbers are indistinguishable. 
How many times does it over count? If all the numbers are the same, I.e. $a=b=c$, then there is just one way.
If two of the numbers are equal, then $2a+c=30$, and there are 14 ways corresponding to $a=1$ to 14. However we would double count $a=b=c$ so there are 13 ways. Each of these 13 ways would lead to 39 ways if the numbers were indistinguishable. 
Now, of the 406 distinguished ways, lets subtract off the $1+39=40$ distinguished ways listed above, giving us 366. These correspond to distinct a, b, c values. Each of these ways are counted 6 times. Hence there are $366/6=61$ undistinguished ways. 
Now we add back the 13 undistinguished from the two same one different, and the 1 undistinguished from three same, and we get $61+13+1=75$
A: If we want the answer to a small problem like this, we can list systematically and count. 
We only count the number of partitions of $30$ into $3$ parts.  The same idea can be used to produce an explicit formula for the number of partitions of $n$ into $3$ parts. For one notices a simple pattern. If we want $4$ parts, a similar idea works. In principle what we use below is a recurrence, from the easy $2$ parts to the less easy $3$ parts. 
We make a list by smallest element in the partition:
Smallest $10$: $1$ way
Smallest $9$: $2$ ways.
Smallest $8$: $4$ ways (next one is $8$ to $11$)
Smallest $7$: $5$ ways
Smallest $6$: $7$ ways (next one $6$ to $12$)
Smallest $5$: $8$ ways
Smallest $4$: $10$ ways.
Smallest $3$: $11$ ways
Smallest $2$: $13$ ways
Smallest $1$: $14$ ways (next one $1$ to $14$)
Note the nice pattern of increase of the numbers. Now add. 
Another way: The following is another direct computational approach. Let $f(n)$ be the number of partitions of $n$ into $3$ parts. The smallest part can be (i) greater than $1$ or (ii) equal to $1$. 
In Case (i), by removing a $1$ from each part, we get a partition of $n-3$, and we get all partitions of $n-3$ in this way. 
To count the Case (ii) possibilities, note that we must partition $n-1$ into two parts. If $n-1$ is odd , this can be done in $(n-2)/2$ ways. If $n-1$ is even it can be done in $(n-1)/2$ ways.
So we have obtained the recurrence
$f(n)=f(n-3)+(n-2)/2$ if $n$ is even, and $f(n)=f(n-3)+(n-1)/2$ if $n$ is odd.
Armed with this recurrence, we can to $n=30$ quickly by $3$'s from the base case $n=3$. 
A: First, I say +1 to Calvin and Andre, nice solutions.
This is covered in the theory of partitions. The use of generating functions is helpful. Not only do they remove the problem of over counting but they generate a sequence of answers which has value that will be seen later. 
The generating function for a partition of a number into 3 positive parts is given by
$$G(x) = \frac{x^3}{(1-x)(1-x^2)(1-x^3)}$$
the theory is given in https://math.berkeley.edu/~mhaiman/math172-spring10/partitions.pdf  number (2)
You only need to determine the coefficient of $x^{30}$. The simplest way to do so is  to take it to Wolfram Alpha and enter "Series[x^3/((1-x)(1-x^2)(1-x^3)),{x,0,30}]" then check the coefficient of $x^{30}$, it will be 75. If you need to, the whole computation can be done by hand using polynomial multiplication or by other methods discussed on this forum. 
You should notice that the polynomial spit out by Alpha contains other coefficients. These are the answers to 3 part partitions summing to 29, 28, 27 ...
Taking the coefficients as a sequence we can now enter 0, 0, 0, 1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19 at the OEIS be sent here http://oeis.org/A069905 and see what is currently known about this sequence of which your problem is a part.
