Question

решить комбинаторику

Answer 1

$1)A_{3} ^{2} =\frac{3!}{(3-2)!}=3!=1*2*3=6 \\\\C_{7}^{5}=\frac{7!}{5!*(7-5)!}=\frac{5!*6*7}{5!*1*2}=21\\\\P_{4}=4!=1*2*3*4=24\\\\\frac{A_{3}^{2}+C_{7}^{5}}{P_{4}}=\frac{6+21}{24}=\frac{27}{24}=1,125\\\\Otvet:\boxed{1,125}$

$2)C_{5}^{3}+A_{5}^{3}=\frac{5!}{3!*(5-3)!}+\frac{5!}{(5-3)!}=\frac{3!*4*5}{3!*1*2}}+\frac{2!*3*4*5}{2!}=10+60=70\\\\Otvet:\boxed{70}\\\\\\3)A_{27}^{2}=\frac{27!}{(27-2)!}=\frac{25!*26*27}{25!}=702\\\\Otvet:\boxed{702}$