Question

Aranjamente de 9 luate cate 3 plus aranjamente de 8 luate cate 4 supra aranjamente de 7 luate cate 3

Answer 1

${A ^ {3} _{9}} + \frac{{A ^ {4} _{8}}}{{A ^ {3} _{7}}} = \frac{9!}{(9 - 3)!} + \frac{ \frac{8!}{(8 - 4)!}}{ \frac{7!}{(7 - 3)!}}$

$= \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9}{6!} + \frac{ \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8}{4!}}{ \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7}{4!}}$

$= \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9}{1 \times 2 \times 3 \times 4 \times 5 \times 6} + \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times7 \times 8 }{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7}$

$= 7 \times 8 \times 9 + 8$

$= 504 + 8$

$= 512$