Find Number of Non negative integer solutions of $3a+2b+c+d=19$
My attempt:
we have $$2b+c+d=19-3a$$ Required solutions is coefficient of $t^{19-3a}$ in
$$( 1-t^2)^{-1}(1-t)^{-1}(1-t)^{-1}=\frac{1}{(1-t)^3(1+t)}$$
By partial fractions we get
$$( 1-t^2)^{-1}(1-t)^{-1}(1-t)^{-1}=\frac{1}{2}\times (1-t)^{-3}+\frac{1}{4} \times (1-t)^{-2}+\frac{1}{4}\times (1-t^2)^{-1}$$
Required coefficient is
$$\frac{1}{2} \sum_{a=0}^{6} \binom{21-3a}{2}+\frac{1}{4} \sum_{a=0}^{6}\binom{20-3a}{1}+\frac{1}{4}(4)=294+\frac{77}{4}+1$$
why i am not getting integer answer?
HINT.-Checking directly is not hard. $$3a+2b=19$$ has only three solutions $(1,8),(3,5),(5,2)$.
When $a=6$,$$2b+c+d=1$$ has two solutions $(0,0,1),(0,1,0)$.
When $a=5$, $$2b+c+d=4$$ has five solutions from which one has to be discarded $(1,1,1)(2,0,0),(0,1,3),(0,2,2),(0,3,1)$ we have till now $$(1,8,0,0),(3,5,0,0),(5,2,0,0)\\(6,0,0,1),(6,0,1,0)\\(5,1,1,1),(5,0,1,3),(5,0,2,2),(5,0,3,1)$$ There are nine solutions as far and we can continue with the remaining values of $a$.