Problems from Combinatorial numbers

Development:

1. $\left(\begin{array}{c}8\\ 3\end{array}\right)={\displaystyle \frac{8!}{3!\cdot (8-3)!}}={\displaystyle \frac{8\cdot 7\cdot 6\cdot 5!}{3!\cdot 5!}}={\displaystyle \frac{8\cdot 7\cdot 6}{3\cdot 2}}=56$

2. $\left(\begin{array}{c}4\\ 1\end{array}\right)=4$

3. $\left(\begin{array}{c}7\\ 2\end{array}\right)={\displaystyle \frac{7!}{2!\cdot (7-2)!}}={\displaystyle \frac{7\cdot 6\cdot 5!}{2!\cdot 5!}}=7\cdot 3=21$

4. $\left(\begin{array}{c}157\\ 0\end{array}\right)=1$

Solution:

1. $56$

2. $4$

3. $21$

4. $1$

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