A6 If $\binom{5}{1}=\frac{5}{1},\binom{5}{2}=\frac{5 \cdot 4}{1 \cdot 2},\binom{5}{3}=\frac{5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3},\binom{5}{4}=\frac{5 \cdot 4 \cdot 3 \cdot 2}{1 \cdot 2 \cdot 3 \cdot 4}$, then $\binom{4}{1}+\binom{4}{3}$ is:
(A) 8
(B) 10
(C) 12
(D) 14
(E) 16
A6. By analogy, $\binom{4}{1}=\frac{4}{1}=4$ and $\binom{4}{3}=\frac{4 \cdot 3 \cdot 2}{1 \cdot 2 \cdot 3}=4$. Therefore, $\binom{4}{1}+\binom{4}{3}=4+4=8$.

## FURTHER PROBLEMS