Repeat Exercise 9 for \(C(7, 5)\), \(C(6, 4)\), and \(C(6, 5)\).

\[C(7, 2) = \binom{7}{2} = \frac{7 \cdot 6}{2 \cdot 1} = 21\]

\[C(6, 4) + C(6, 5) = \binom{6}{4} + \binom{6}{5} = \frac{6 \cdot 5 \cdot 4 \cdot 3}{4 \cdot 3 \cdot 2 \cdot 1} + \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 15 + 6 = 21\]