SYMBOL NEWTONA - ZADANIA
Zadanie 1
Oblicz : a) \(\begin{pmatrix}4\\0\end{pmatrix}\) b) \(\begin{pmatrix}4\\1\end{pmatrix}\) c) \(\begin{pmatrix}4\\2\end{pmatrix}\)
d) \(\begin{pmatrix}4\\3\end{pmatrix}\) e) \(\begin{pmatrix}4\\4\end{pmatrix}\) Rozwiązanie
W rozwiązywaniu zadań korzystamy ze wzoru : \(\begin{pmatrix}n\\k\end{pmatrix}=\large\frac{n!}{k!\left(n-k\right)!}\)
a) \(\displaystyle \begin{pmatrix}4\\0\end{pmatrix}=\frac{4!}{0!\cdot\left(4-0\right)!}=\frac{4!}{1\cdot4!}=1\)
Należy pamiętać, że : \(0!=1\)
b) \(\displaystyle\begin{pmatrix}4\\1\end{pmatrix}=\frac{4!}{1!\cdot(4-1)!}=\frac{4!}{1\cdot3!}=\frac{\not3!\cdot4}{\not3!}=4\)
c) \(\displaystyle\begin{pmatrix}4\\2\end{pmatrix}=\frac{4!}{2!\cdot\left(4-2\right)!}=
\frac{4!}{2!\cdot2!}=\frac{\not{2!}\cdot3\cdot\not4^2}{\not{2!}\cdot1\cdot{\not2}_1}=6\)
d) \(\displaystyle\begin{pmatrix}4\\3\end{pmatrix}=
\frac{4!}{3!\cdot\left(4-3\right)!}=
\frac{4!}{3!\cdot1!}=\frac{\not3!\cdot4}{\not3!}=4\)
e) \(\displaystyle\begin{pmatrix}4\\4\end{pmatrix}=
\frac{4!}{4!\cdot\left(4-4\right)!}=\frac{\not4!}{\not4!\cdot0!}=\frac11=1\)