Question

Number of non negative integral solutions of $A+B+C+D=12$
where $A,B>0$ and $0<D<5$ is

Answer 1

Number of solutions for $A+B+C+D=12$
where $A,B\text{and}D>0$ is $A+B+C+D=9$$\Rightarrow {}^{9+4-1}{C}_{4-1}={}^{12}{C}_3=220...\left(1\right)$

Now from eqn $1$ subtrating those cases where $D\ge 5$ then $A+B+C+D=5$$\Rightarrow {}^{5+4-1}{C}_{4-1}={}^8{C}_3=56$
Required value is $220-56=164$