[Combinatorics]
Problem: The zeros of $f(x)=x^{6}+2 x^{5}+3 x^{4}+5 x^{3}+8 x^{2}+13 x+21$ are distinct complex numbers. Compute the average value of $A+B C+D E F$ over all possible permutations $(A, B, C, D, E, F)$ of these six numbers.

Solution: There are $6 !=720$ permutations of the zeros, so the average value is the sum, $S$, divided by 720. Setting any particular zero as $A$ leaves $5 !=120$ ways to permute the other five zeros, so over the 720 permutations, each zero occupies the $A$ position 120 times. Similarly, fixing any ordered pair $(B, C)$ of zeros allows $4 !=24$ permutations of the other four zeros, and $B C=C B$ means that each value of $B C$ occurs 48 times. Finally, fixing any ordered triple $(D, E, F)$ allows $3 !=6$ permutations of the other variables, and there are $3 !=6$ equivalent arrangements within each product $D E F$, so that the product of any three zeros occurs 36 times within the sum. Let $S_{1}=A+B+C+D+E+F$ (i.e., the sum of the zeros taken singly), $S_{2}=A B+A C+\cdots+A F+B C+\cdots+E F$ (i.e., the sum of the zeros taken two at a time), and $S_{3}=A B C+A B D+\cdots+D E F$ be the sum of the zeros three at a time. Then $S=120 S_{1}+48 S_{2}+36 S_{3}$. Using the sums and products of roots formulas, $S_{1}=-2 / 1=-2$, $S_{2}=3 / 1=3$, and $S_{3}=-5 / 1=-5$. Thus $S=120(-2)+48(3)+36(-5)=-276$. The average value is thus $-\frac{276}{720}=-\frac{\mathbf{2 3}}{\mathbf{6 0}}$.

Final answer: $-\frac{23}{60}$