Problems from Factorial and combinatorial numbers

Development:

a) $6!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720$

b) $${\displaystyle (\genfrac{}{}{0ex}{}{7}{2})=\frac{7!}{2!\cdot (7-2)!}={\displaystyle \frac{7!}{2!\cdot 5!}}={\displaystyle \frac{7\cdot 6\cdot \text{⧸}5\cdot \text{⧸}4\cdot \text{⧸}3\cdot \text{⧸}2\cdot 1}{2\cdot 1\cdot \text{⧸}5\cdot \text{⧸}4\cdot \text{⧸}3\cdot \text{⧸}2\cdot 1}}={\displaystyle \frac{7\cdot 6}{2}}=21}$$

Solution:

a) $6!=720$

b) ${\displaystyle (\genfrac{}{}{0ex}{}{7}{2})=21}$

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