NOTE: I know there are similar questions to this, but the ones on this website are much more complex, and I'd like to get a basic understanding before moving on to them. Please do not mark this as a duplicate.
On a practice quiz:
Use inc-exc to determine the number of solutions to $a+b+c+d = 15$ with $0 \leq a,b,c,d \leq 6$
Here's what I have so far:
Let $A_1$ be the solution to $a+b+c+d=15$ with $a\geq 7, 0\leq b,c,d$
Let $A_2$ be the solution to $a+b+c+d=15$ with $b\geq 7, 0\leq a,c,d$
Let $A_3$ be the solution to $a+b+c+d=15$ with $c\geq 7, 0\leq a,b,d$
Let $A_4$ be the solution to $a+b+c+d=15$ with $d\geq 7, 0\leq a,b,c$
$|A_1 \cup A_2 \cup A_3 \cup A_4| = C(4,1) \cdot C((4+4-1), 4) - \dotsb $
And this is where I get stuck. I found my first term using the logic that I want to count $|A_1|+|A_2|+|A_3|+|A_4|$. How do I continue this to find the rest of the terms? I know what to do after I find the number of non-solutions, just subtract it from the total non-restricted solutions...but I can't figure out how to find the number of non-solutions!
Any help would be greatly appreciated, thanks.
Generating Functions
Look at the coefficient of $x^{15}$ in $\left(1+x+x^2+x^3+\dots+x^6\right)^4$: $$ \begin{align} \left(\frac{1-x^7}{1-x}\right)^4 &=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty(-1)^k\binom{-4}{k}x^k\\ &=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty\binom{k+3}{k}x^k \end{align} $$ which is $$ \sum_{j=0}^2(-1)^j\binom{4}{j}\binom{18-7j}{3}=180 $$
Inclusion-Exclusion
Without restriction on the size of the terms, using the standard $\mid$ and $\circ$ argument ($15$ $\circ$s and $3$ $\mid$s), there are $\binom{15+3}{3}$ ways to choose 4 non-negative integers that sum to $15$. $$ \text{one sum for each arrangement}\\ 2+4+6+3=\circ\,\circ\mid \circ\circ\circ\,\circ\mid \circ\circ\circ\circ\circ\,\circ\mid \circ\circ\circ $$ Now let's count how many ways there are to have terms greater than $6$. There are $\binom{4}{1}$ ways to choose which $1$ term should be greater than $6$. To count the number of sums with $1$ term at least $7$, that would be $\binom{15-7+3}{3}$. $$ \text{consider the red group atomic}\\ 2+8+4+1=\circ\,\circ\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ\circ}\,\circ\mid\circ\circ\circ\,\circ\mid\circ $$ There are $\binom{4}{2}$ ways to choose which $2$ terms should be greater than $6$. To count the number of sums with $2$ terms at least $7$, that would be $\binom{15-14+3}{3}$. $$ 7+0+7+1=\color{#C00000}{\circ\circ\circ\circ\circ\circ\,\circ}\mid\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ\circ}\mid\circ $$ There is no way for $3$ terms to be greater than $6$. Inclusion-Exclusion says there are $$ \binom{18}{3}-\binom{4}{1}\binom{11}{3}+\binom{4}{2}\binom{4}{3}=180 $$ ways for $4$ terms to sum to $15$ with each term at most $6$.