Question

Quantas comissões com 2 membros podemos formar com 4 alunos

Answer 1

Olá !

Resolução :

$\mathsf{ Cn,p =\dfrac{n!}{p!(n - p)!}} \\\\\\ \mathsf{ C4,2 =\dfrac{4!}{2!(4 - 2)!}} \\\\\\ \mathsf{ C4,2 =\dfrac{4!}{2!2!}} \\\\\\ \mathsf{ C4,2 =\dfrac{24}{4}} \\\\\\ \mathsf{ C4,2 = 6}$

Resposta :

$\mathsf{6~~Comissoes}$