Inspired by Aviel Livay

Substituting $x = 51-a, y = 51-b, z = 51-c$ gives $x+y+z = 52$ with the conditions $1 \le x,y,z \le 50$ . However, any integral solution to $x+y+z = 52$ with $x,y,z \ge 1$ automatically satisfies $x,y,z \le 50$ (otherwise, $x+y+z \ge 51+1+1 > 52$ ), so we can effectively ignore that condition. Now this is a standard problem: the number of 3-tuples of positive integers whose sum is 52, which is given by the formula $\binom{52-1}{3-1} = \boxed{1275}$ . (The proof is easy: consider 52 1's in a line and place two separators between some pairs of the 1's (must go between different pairs). 51 places to put 2 separators.)

Put $a=50-k_1 ; b=50-k_2 ; c=50-k_3$ as $k_1,k_2,k_3$ are non-negative integers.

$150-(k_1+k_2+k_3)=101 \implies k_1+k_2+k_3=49$

The non-negative integers solutions are $\binom{49+3-1}{3-1}=\binom{51}{2}=1275$

Now the important point is that every solutions from above is allowable because every $k_i$ vary from $0 \to 49$ that do not hamper the given limit in the question given.

Therefore,total solutions are $1275$

Let $f(n)$ denote the number of ordered triplets $(a,b,c)$ of integers such that $1 \leq a,b,c \leq 50$ and $a+b+c=n$ . Then $\sum_{n=0}^\infty f(n) x^n = (x+x^2+\cdots + x^{50})^3 = \left(\frac{x-x^{51}}{1-x}\right)^3 = \frac{x^3 - 3x^{53} + \cdots}{(1-x)^3} = (x^3 - 3x^{53} + \cdots)\sum_{n=0}^\infty \binom{n+2}{2} x^n$

Selecting only the coefficient of $x^{101}$ , we have $f(101) = \binom{100}{2} - 3\binom{50}{2} = \boxed{1275}.$

We can use generating functions. For each a,b,c the generating function is $\left(\frac{x\left(1-x^{50}\right)}{1-x}\right)$ . Multiply them we'll get $\left(\frac{x\left(1-x^{50}\right)}{1-x}\right)^3$ . We're trying to find the 101 coefficient. Since it's being multiplied by $x^3$ , we can simply get rid of the $x^3$ and subtract the 101 by 3 resulting in 98, $\left[x^{98}\right]\frac{\left(1-x^{50}\right)^3}{\left(1-x\right)^3}$ . We can can think of this as what will equal $x^{98}$ by multiplying the numerator and denomanator. One way by ignoring the $x^{50}$ we will get $\operatorname{nCr}\left(-3,98\right)$ which by using the general binomial coefficient we'll get $\operatorname{nCr}\left(100,98\right)$ which equals 4950. Then we have to account where we use 1 $x^{50}$ . This will be $-\operatorname{nCr}\left(3,1\right)*\operatorname{nCr}\left(50,48\right)$ . It's negative since we coef. next to $x^{50}$ is negative and $\operatorname{nCr}\left(50,48\right)$ since we take 1 $x^{50}$ and what plus 50 equals 98? 40. So then we'll use the negative binomial coef. again and get $\operatorname{nCr}\left(50,48\right)$ . Adding $\operatorname{nCr}\left(100,98\right)$ with $-\operatorname{nCr}\left(3,1\right)*\operatorname{nCr}\left(50,48\right)$ results in 1275 .