Too many solutions!

Here we can see that we have $x+ y +z = 102$ for all $+ve$ integral solutions. We know that if we need to do partitions for $n$ distinct things into $3$ divisions and they are non negetive integers then think about lets say there are $102$ balls and we have to divide in $3$ divisions . then we need $2$ sticks for this division but they can be placed any where so between two sticks we may also get $0$ balls. So we see for $102$ balls is we place \(2 ) sticks of partitions then $103C2$ are ways to arrange them will be our answer similarly for $n$ balls and $r$ partions our formulae will be $n+r-1Cr-1$ , But $(x,y,z)$ is $+ve$ integral then $x>0$ , $y>0$ and $z>0$ or $x>=1$ and $y>=1$ and $z>=1$ or now $x-1=x'$ , $y-1=y'$ and $z-1 = z'$ . Hence Now putting $x' + y'+ z' = 99$ . From the previous formulae $x'$ , $y'$ and $z'$ are brought for non negetive. Hence our number of triplets $(x,y,z)$ for only +ve integers will be $99+3-1+C3-1$ = $101C2$ .