Question

Quantas são as combinações de 8 elementos tomados 2 a 2?

Answer 1

$C_{n, p} = \frac{n!}{p!(n-p)!} \\C_{8, 2} = \frac{8!}{2!(8-2)!} \\C_{8, 2} = \frac{8!}{2!6!} \\C_{8, 2} = \frac{8 * 7 * 6!}{2!6!} \\C_{8, 2} = \frac{8 * 7}{2 * 1} \\C_{8, 2} = 4 * 7\\C_{8, 2} = 28$