Find the number of natural number solutions of $a+2b+c=100$ 
Find the number of natural number solutions of $a+2b+c=100$

I remember something like stars and bars if the equation I change to $a+b_{1}+b_{2}+c=100$
then i get $\dbinom{99}{3}$ ways.
If the equation is like $a_{1}+a_{2}+a_{3}+a_{4}=m$ I can use $\dbinom{m-1}{3}$ ways
how ever I am not familier with variation in the problem.
I look for a short and simple way.
I have studied maths upto $12$th grade. thanks.
 A: I will assume that by the natural numbers that you mean the positive integers (as opposed to the nonnegative integers).
Observe that $a + c = 100 - 2b$ is an even number.  Thus, $a$ and $c$ must have the same parity.
Case 1:  The numbers $a$ and $c$ are both even.  Let $a = 2u$; let $c = 2v$.  Then $u, v \in \mathbb{N}$.  Moreover,
\begin{align*}
a + 2b + c & = 100\\
2u + 2b + 2v & = 100\\
u + b + v & = 50 \tag{1}
\end{align*}
which is an equation in positive integers. The number of solutions of equation 1 is the number of ways we can place two addition signs in the $49$ spaces between consecutive ones in a row of $50$ ones, which is $\binom{49}{2}$.  
Case 2:  The numbers $a$ and $c$ are both odd.  Let $a = 2s - 1$; let $b = 2t - 1$.  Then $s, t \in \mathbb{N}$.  Moreover,
\begin{align*}
a + 2b + c & = 100\\
2s - 1 + 2b + 2t - 1 & = 100\\
2s + 2b + 2t & = 102\\
s + b + t & = 51 \tag{2}
\end{align*}
which is an equation in positive integers.  The number of solutions of equation 2 is the number of ways two addition signs can be placed in the $50$ spaces that appear between consecutive ones in a row of $51$ ones, which is $\binom{50}{2}$.
Thus, the number of solutions of the equation $a + 2b + c = 100$ in the positive integers is 
$$\binom{49}{2} + \binom{50}{2} = 1176 + 1225 = 2401$$
A: I assume that $a,b,c\ge 1$.
You can use the method you write for 
$$a+c=100-2b$$
where $b=1,2,\cdots,49$. 
Then, the answer is
$$\sum_{b=1}^{49}\binom{99-2b}{1}=99\times 49-2\cdot\frac{49\cdot 50}{2}=49(99-50)=49^2=\color{red}{2401}.$$
