Error in calculation of number of ways to put $20$ identical balls in $4$ labelled boxes if each box contains at most $18$ balls The number of ways to put $20$ identical balls in $4$ labelled boxes in such a way that each box contains at most $18$ balls is,  
$$(a)~~\binom{24}{4}-16~~(b)~~\binom{24}{4}-10~~(c)~~\binom{23}{3}-16~~(d)~~\binom{23}{3}-12~~(e)~~\binom{24}{4}-12$$
My solution:
Answer - Stars bars All balls can go in each box - $4C1\cdot$ Stars bars with $2$ balls and $3$ boxes (Fill $1$ box with $18$ balls leaves $2$ balls and $3$ boxes. AND $4C1$ ways of selecting full box ) 
$$\begin{align}
&= (23 C 3) - (4 C 1) \cdot ( 4 C 2)\\
&= (23 C 3) - 4 \cdot 6\\
&= (23 C 3) - 24
\end{align}$$
Where did I go wrong? 

We need the coefficient of $t^{20}$ in the expansion $(1+t+\cdots +t^{18})^4$.
We have
\begin{align*}
(1+t+\cdots +t^{18})^4 &= \left(\dfrac{1-t^{19}}{1-t}\right)^4 \\
&= (1-t^{19})^4(1-t)^{-4} \\
&= (1-4t^{19} + \cdots )\sum_{n=0}^\infty \binom{n+3}{n}t^n
\end{align*}
and hence equals $\binom{23}{20} - 4\binom{4}{1} = \binom{23}{3} - 16$.
