<s>[INST] Find the number of solutions in nonnegative integers of the (strict) inequality $a+b+c+d+e<11$ when $a$ is at most $6$. Find the number of solutions in nonnegative integers of the (strict) inequality $a+b+c+d+e<11$.
We can see that $a+b+c+d+e \leq10$ and so there are $10+5 \choose 5$ solutions.
Now, what if $a$ is at most $6$? That is, $a\leq6$. This means $a\notin\{7,8,9,10\}$. I decided to use the subtraction principle. I let $a=7,8,9,10$ (four separate cases) and came to ${15 \choose 5}-{7 \choose 4}-{6 \choose 4}-{5 \choose 4}-{4 \choose 4}=3003-56=2947$ which is correct. Is there a better way to do this? Is there a quicker way to do this? The textbook solution is ${15 \choose 5}-{8 \choose 5}$ and I wonder how they came across that exact form. Any solutions or hints are greatly appreciated.
 [/INST] The shortcut is that $a + b + c + d + e \leq 10$, where $a \geq 7$ bijects to $a + b + c + d + e \leq 3$, where $a,b,c,d,e$ are non-negative integers.
Therefore, the number of solutions to deduct is
$$\binom{7}{4} + \binom{6}{4} + \binom{5}{4} + \binom{4}{4} = \binom{8}{5}.$$
</s>