Number of Non negative integral solutions of $x+2y+3z+4w=12$ Find Number of positive integral solutions of $x+2y+3z+4w=12$
I tried to split the problem in to cases :
we have $2z+2y+4w$ is always  Even and $12$ is Even number.
Hence $x+z$ should be even number which implies
Case $1.$ $x$ is Even and $z$ is even
Letting $x=2p$ and $z=2q$ we get
$p+y+3q+2w=6$
Now Let $q=0$ we have
$$p+y+2w=6$$
$$p+y=6-2w$$
Number of solutions by stars and bars is given by
$$\sum_{w=0}^{3} 7-2w=16$$
Again letting $q=1$ we have to repeat same process.
This method i feel is so lengthy. Any other way?

Minimum of $x+2y+3z=6\Rightarrow 4w=4$ since $8+6\gt12$ then $w=1$ necessarily.
Minimum of $x+2y+4w=7\Rightarrow 3z=3$ since $7+6\gt12$ then $z=1$ necessarily.
It follows $$x+2y=12-3-4=5$$ The only solutions of this equation are $(x,y)=(1,2),(3,1)$.
Thus there are only two solutions
$$(x,y,z,w)=(1,2,1,1),(3,1,1,1)$$
