Jeżeli k≤n, to wszystkich k-elementowy kombinacji zbioru n-elementowego jest: $C_n^k=\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}$

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a)

$\left(\begin{array}{c}4\\ 0\end{array}\right)=\frac{4!}{0!\cdot \left(4-0\right)!}=\frac{4!}{1\cdot 4!}=1$  

$\left(\begin{array}{c}4\\ 1\end{array}\right)=\frac{4!}{1!\left(4-1\right)!}=\frac{4\cdot 3!}{1\cdot 3!}=4$

$\left(\begin{array}{c}4\\ 2\end{array}\right)=\frac{4!}{2!\cdot \left(4-2\right)!}=\frac{4\cdot 3\cdot 2!}{2\cdot 2!}=2\cdot 3=6$

$\left(\begin{array}{c}4\\ 3\end{array}\right)=\frac{4!}{3!\left(4-3\right)!}=\frac{4\cdot 3!}{3!\cdot 1!}=\frac{4}{1}=4$

$\left(\begin{array}{c}4\\ 4\end{array}\right)=\frac{4!}{4!\left(4-4\right)!}=\frac{1}{0!}=\frac{1}{1}=1$      

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b)

$\left(\begin{array}{c}5\\ 0\end{array}\right)=\frac{5!}{0!\cdot \left(5-0\right)!}=\frac{5!}{1\cdot 5!}=1$

$\left(\begin{array}{c}5\\ 1\end{array}\right)=\frac{5!}{1!\left(5-1\right)!}=\frac{5\cdot 4!}{1\cdot 4!}=5$

$\left(\begin{array}{c}5\\ 2\end{array}\right)=\frac{5!}{2!\cdot \left(5-2\right)!}=\frac{5\cdot 4\cdot 3!}{2\cdot 3!}=5\cdot 2=10$

$\left(\begin{array}{c}5\\ 3\end{array}\right)=\frac{5!}{3!\left(5-3\right)!}=\frac{5\cdot 4\cdot 3!}{3!\cdot 2!}=\frac{20}{2}=10$

$\left(\begin{array}{c}5\\ 4\end{array}\right)=\frac{5!}{4!\left(5-4\right)!}=\frac{5\cdot 4!}{4!\cdot 1!}=\frac{5}{1}=5$    

$\left(\begin{array}{c}5\\ 5\end{array}\right)=\frac{5!}{5!\left(5-5\right)!}=\frac{1}{0!}=\frac{1}{1}=1$