Korzystamy ze wzoru na symbol Newtona: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ oraz definicji silni: $n! = 1 \cdot 2 \cdot ... \cdot n$. a) Obliczamy $\binom{7}{3}$.\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} = \frac{4! \cdot 5 \cdot 6 \cdot 7}{3! \cdot 4!} = \frac{5 \cdot 6 \cdot 7}{6} = 35 \]b) Obliczamy $3 \cdot \binom{6}{4}$. Korzystamy z własności $\binom{n}{k} = \binom{n}{n-k}$, więc $\binom{6}{4} = \binom{6}{2}$.\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} = \frac{4! \cdot 5 \cdot 6}{2 \cdot 4!} = \frac{30}{2} = 15 \]Zatem:\[ 3 \cdot \binom{6}{4} = 3 \cdot 15 = 45 \]c) Obliczamy $\binom{9}{2} - \binom{6}{3}$. Najpierw poszczególne symbole Newtona:\[ \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2! \cdot 7!} = \frac{7! \cdot 8 \cdot 9}{2 \cdot 7!} = \frac{72}{2} = 36 \]\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!} = \frac{3! \cdot 4 \cdot 5 \cdot 6}{3! \cdot 3!} = \frac{120}{6} = 20 \]Ostatecznie:\[ \binom{9}{2} - \binom{6}{3} = 36 - 20 = 16 \]d) Obliczamy $\binom{11}{10} \cdot \binom{8}{8}$. Korzystamy z własności $\binom{n}{n-1} = n$ oraz $\binom{n}{n} = 1$.\[ \binom{11}{10} = 11 \]\[ \binom{8}{8} = 1 \]Zatem:\[ \binom{11}{10} \cdot \binom{8}{8} = 11 \cdot 1 = 11 \]e) Obliczamy $\binom{7}{4} \cdot 2!$.\[ \binom{7}{4} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} = \frac{4! \cdot 5 \cdot 6 \cdot 7}{3! \cdot 4!} = \frac{210}{6} = 35 \]\[ 2! = 1 \cdot 2 = 2 \]Zatem:\[ \binom{7}{4} \cdot 2! = 35 \cdot 2 = 70 \]f) Obliczamy $\binom{8}{3} - 5!$.\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} = \frac{5! \cdot 6 \cdot 7 \cdot 8}{3! \cdot 5!} = \frac{336}{6} = 56 \]\[ 5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 \]Zatem:\[ \binom{8}{3} - 5! = 56 - 120 = -64 \]