@Raja Raval solved it Very nicely. I am trying one more approach here.
We have to find
$$x _ { 1 } + x _ { 2 } + \ldots + x _ { k } = n , \text { where } x _ { i } \in \{ 0,1,2,3 , \dots \}$$
if all x1,x2,..xn belongs to {0,1,2,3....}
answer will be $\left( \begin{array} { c } { n + k - 1 } \\ { k -1 } \end{array} \right) = \left( \begin{array} { c } { 100 + 4 - 1 } \\ { 4 - 1 } \end{array} \right)$
why because it is same as filling n holes from k type of bagels replacement (Watch Above Video !!!)
but our x1 >= 1 and x2 >= 2 so if we write y1 = x1 - 1 and y2 = x2 - 2
then we are left with n = 100 - (1+2) = 97 ( why because .... we have already chosen 1 object of x1 and 2 objects of x2)
our problem boils down to
$$y _ { 1 } + y_ { 2 } + \ldots + x _ { k } = n , \text { where } x _ { i } \in \{ 0,1,2,3 , \dots \}$$
all y1,y2,x2..xk belongs to {0,1,2,3....}
So the Answer is
$$ \left( \begin{array} { c } { n + k - 1 } \\ { k -1 } \end{array} \right) = \left( \begin{array} { c } { 97 + 4 - 1 } \\ { 4 - 1 } \end{array} \right) = \left( \begin{array} { c } { 100 } \\ { 3 } \end{array} \right) = \left( \begin{array} { c } { 100 } \\ { 97 } \end{array} \right) $$