We have an even number of pens, $n=2k$ pens, and $3$ boxes called $A$, $B$ and $C$. In box $A$ we need to have a minimum of $2$ pens and in box $C$ we have a maximum of $\frac{n}{2}=k$ pens. How many possible ways can we divide up the pens?
I "tried" to solve this problem but I can't even decide where to begin. How can I calculate the possible number of arrangements?
Let $x_A$, $x_B$, $x_C$ be the number of pens in each box. Then $$x_A+x_B+x_C=2k.$$ with $x_A\geq 2$, $x_B\geq 0$ and $x_C\leq k=n/2$.
Let $y_A=x_A-2\geq 0$ then, by using stars-and -bars, the number of non-negative solutions of $$y_A+x_B+x_C=2k-2$$ is $n_1=\binom{2k}{2}$ .
Similarly let $y_C=x_C-(k+1)\geq 0$ (so the second condition is NOT satisfied) then count the number $n_2$ of non-negative solutions of $$y_A+x_B+y_C=2k-2-(k+1).$$ Can you take it from here?