<s>[INST] Solution to equation using permutation I am hoping I am on the right track for my homework in Discrete Math. 

How many solutions does the equation $$a + b + c + d + e + f = 30$$ have if each variable must be a non-negative integer and $a ≤ 3, b ≤ 7$ and $d ≥ 8$?

I started my solution using the formula: ${m+n-1 \choose n-1}$ where m is the number of non-distinct objects and n is the number of distinct objects.
d ≥ 8 where ${22+9-1 \choose 9-1}$ = ${30 \choose 8}$
Then, a ≤ 3 which can be represented as a ≥ 4 
${17+6-1 \choose 6-1}$ = ${21 \choose 5}$
Finally, b ≤ 7 which can be represented as b ≥ 8
${9+6-1 \choose 6-1}$ = ${14 \choose 5}$
During this process we have removed solutions for a ≤ 3 and b ≤ 7 twice so we have to add the solution back in once. We removed 25 - (3+7) = 15 so we must add back ${15+6-1 \choose 6-1}$ = ${20 \choose 5}$
Therefore, our solution is as follows:
${30 \choose 8}$ - ${21 \choose 5}$ - ${14 \choose 5}$+ ${20 \choose 5}$
 [/INST] The number of all 6-couples in our universe, which is set of all solutions of the equation $a+b+c+d'+e+f =22$ where $d'=d-8\geq 0$ is $${27\choose 5}$$
Let $A$ be a set of all 6-couples where $a\geq 4$ and $B$ be a set of all 6-couples where $b\geq 8$. 
So $$|A| = {23\choose 5}\;\;\;{\rm and}\;\;\;|B| = {19\choose 5}$$
Since $$|A\cap B| = {15\choose 5}$$ the finaly answer is (using formula of inclusion and exclusion) $$ {27\choose 5}- {23\choose 5}- {19\choose 5}+ {15\choose 5}$$
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