I am looking to get some help with the following problem, I am not sure if my workings are correct but I have given it a good shot, thanks!
How many ways are there for a child to take $11$ pieces of candy with $4$ types of candy if they child does not take exactly two pieces of candy of any type?
My idea was to see it up and think of it in the way that there are 11 pieces balls, that have to go into 4 bins, but you cant take from the same bin twice.
So I would use the negation $N(a_1', a_2', a_3', a_4')$
Then the total number of ways would be $N(u) = \binom{11+4-1}{11}$
Then for each next term take $5$ off so
$$N(a_1', a_2', a_3', a_4') = \binom{11+4-1}{11}-\binom{4}{1}\binom{6+4-1}{6} + \binom{4}{2}\binom{1+4-1}{1}-outofballs$$
Assuming I understand your notation correctly, you incorrectly have $N(a_1)=\binom{6+4-1}6$, when it should be $$N(a_1)=\binom{9+3-1}{9}.$$ If the child took $2$ of the first type of candy, there are $\binom{9+3-1}{9}$ ways to choose $9$ candies from the other $3$ types of candy.
Similarly, $N(a_1,a_2) = \binom{7+2-1}7$ and $N(a_1,a_2,a_3) = \binom{5+1-1}5$, while $N(a_1,a_2,a_3,a_4) = 0$.