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:
$\begin{align}
1+1+28 \\
10+10+10 \\
1+2+27 \\
…
\end{align}$
Best Answer
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 $61+13+1=75$