Twierdzenie $\left(\begin{array}{c}n\\ k\end{array}\right)=\left(\begin{array}{c}n\\ n-k\end{array}\right),\text{gdzie}n\in \mathbb{N},k\in \mathbb{N},k\le n$     $\left(\begin{array}{c}n\\ k\end{array}\right)=\left(\begin{array}{c}n-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-1\\ k\end{array}\right),\text{gdzie}n\in \mathbb{N},k\in \mathbb{N},k\le n-1$

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a)

Korzystając z własności (1) symbolu Newtona dostajemy 

$\left(\begin{array}{c}18\\ 16\end{array}\right)=\left(\begin{array}{c}18\\ 18-16\end{array}\right)=\left(\begin{array}{c}18\\ 2\end{array}\right)$   

Zatem 

$\left(\begin{array}{c}18\\ 2\end{array}\right)-\left(\begin{array}{c}18\\ 16\end{array}\right)=\left(\begin{array}{c}18\\ 2\end{array}\right)-\left(\begin{array}{c}18\\ 2\end{array}\right)=0$  

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b)

Korzystając z własności (2) symbolu Newtona dostajemy 

$\left(\begin{array}{c}23\\ 21\end{array}\right)+\left(\begin{array}{c}23\\ 22\end{array}\right)=\left(\begin{array}{c}24\\ 22\end{array}\right)=\frac{24!}{22!\cdot \left(24-22\right)!}=\frac{{\sout{22!}}^1\cdot 23\cdot 24}{{\sout{22!}}_1\cdot 2!}=276$  

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c)

Korzystając z własności (2) symbolu Newtona dostajemy 

$\left(\begin{array}{c}25\\ 23\end{array}\right)+\left(\begin{array}{c}25\\ 24\end{array}\right)=\left(\begin{array}{c}26\\ 24\end{array}\right)=\frac{26!}{24!\cdot \left(26-24\right)!}=\frac{{\sout{24!}}^1\cdot 25\cdot 26}{{\sout{24!}}_1\cdot 2!}=325$  

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d)

Korzystając z własności (1) symbolu Newtona dostajemy 

$\left(\begin{array}{c}6\\ 2\end{array}\right)=\left(\begin{array}{c}6\\ 6-2\end{array}\right)=\left(\begin{array}{c}6\\ 4\end{array}\right)=\frac{6!}{4!\cdot \left(6-4\right)!}=\frac{4!\cdot 5\cdot 6}{4!\cdot 2!}=15$  

Zatem mamy 

$\frac{3\cdot \left[\left(\begin{array}{c}6\\ 2\end{array}\right)\left(\begin{array}{c}15\\ 13\end{array}\right)+\left(\begin{array}{c}6\\ 4\end{array}\right)\cdot \left(\begin{array}{c}15\\ 14\end{array}\right)\right]}{\left(\begin{array}{c}16\\ 2\end{array}\right)}=\frac{3\cdot \left[15\cdot \left(\begin{array}{c}15\\ 13\end{array}\right)+15\cdot \left(\begin{array}{c}15\\ 14\end{array}\right)\right]}{\left(\begin{array}{c}16\\ 2\end{array}\right)}=$  

$=\frac{45\cdot \left[\left(\begin{array}{c}15\\ 13\end{array}\right)+\left(\begin{array}{c}15\\ 14\end{array}\right)\right]}{\left(\begin{array}{c}16\\ 2\end{array}\right)}\stackrel{\text{(2)}}{=}\frac{45\cdot \left(\begin{array}{c}16\\ 14\end{array}\right)}{\left(\begin{array}{c}16\\ 2\end{array}\right)}\stackrel{\text{(1)}}{=}\frac{45\cdot \left(\begin{array}{c}16\\ 16-14\end{array}\right)}{\left(\begin{array}{c}16\\ 2\end{array}\right)}=\frac{45\cdot \left(\begin{array}{c}16\\ 2\end{array}\right)}{\left(\begin{array}{c}16\\ 2\end{array}\right)}=45$  

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e)

Korzystając z własności (1) symbolu Newtona dostajemy 

$\left(\begin{array}{c}24\\ 19\end{array}\right)=\left(\begin{array}{c}24\\ 24-19\end{array}\right)=\left(\begin{array}{c}24\\ 5\end{array}\right)$

Zatem mamy 

$\frac{\left(\begin{array}{c}24\\ 5\end{array}\right)}{\left(\begin{array}{c}24\\ 19\end{array}\right)}=\frac{\left(\begin{array}{c}24\\ 5\end{array}\right)}{\left(\begin{array}{c}24\\ 5\end{array}\right)}=1$  

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f)

Korzystając z własności (1) symbolu Newtona dostajemy 

•   $\left(\begin{array}{c}5\\ 4\end{array}\right)=\left(\begin{array}{c}5\\ 5-4\end{array}\right)=\left(\begin{array}{c}5\\ 1\end{array}\right)=5$  
   
• $\left(\begin{array}{c}5\\ 3\end{array}\right)=\left(\begin{array}{c}5\\ 5-3\end{array}\right)=\left(\begin{array}{c}5\\ 2\end{array}\right)=\left(\frac{5!}{2!\cdot \left(5-2\right)!}\right)=\frac{5!}{2!\cdot 3!}=\frac{3!\cdot 4\cdot 5}{2\cdot 3!}=10$  

Zatem mamy 

$\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)\cdot \left(\begin{array}{c}17\\ 14\end{array}\right)+\left(\begin{array}{c}17\\ 15\end{array}\right)\cdot \left(\begin{array}{c}5\\ 4\end{array}\right)}{\left(\begin{array}{c}5\\ 2\end{array}\right)\cdot \left(\begin{array}{c}17\\ 2\end{array}\right)+\left(\begin{array}{c}5\\ 3\end{array}\right)\cdot \left(\begin{array}{c}17\\ 3\end{array}\right)}=\frac{5\cdot \left(\begin{array}{c}17\\ 14\end{array}\right)+\left(\begin{array}{c}17\\ 15\end{array}\right)\cdot 5}{10\cdot \left(\begin{array}{c}17\\ 2\end{array}\right)+10\cdot \left(\begin{array}{c}17\\ 3\end{array}\right)}=$  

$=\frac{{\sout{5}}^1\cdot \left[\left(\begin{array}{c}17\\ 14\end{array}\right)+\left(\begin{array}{c}17\\ 15\end{array}\right)\right]}{{\sout{10}}_2\cdot \left[\left(\begin{array}{c}17\\ 2\end{array}\right)+\left(\begin{array}{c}17\\ 3\end{array}\right)\right]}\stackrel{\text{(2)}}{=}\frac{\left(\begin{array}{c}18\\ 15\end{array}\right)}{2\cdot \left(\begin{array}{c}18\\ 3\end{array}\right)}\stackrel{\text{(1)}}{=}\frac{\left(\begin{array}{c}18\\ 18-15\end{array}\right)}{2\cdot \left(\begin{array}{c}18\\ 3\end{array}\right)}=\frac{\left(\begin{array}{c}18\\ 3\end{array}\right)}{2\cdot \left(\begin{array}{c}18\\ 3\end{array}\right)}=\frac{1}{2}$