Tehtävä 143

a)
$\frac{6!}{4!}=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1}=\frac{6\cdot5}{1}=\frac{30}{1}=30$
b)
$\frac{6!}{\left(6-3\right)!}=\frac{6!}{3!}=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}=\frac{120}{1}=120$
c)
$\frac{7!}{2!5!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{7\cdot6}{2\cdot1}=\frac{42^{\left(2\right)}}{2}=\frac{21}{1}=21$