Question

utilizando a fórmula de combinação, calcule C11,4 e C7,3 ?
MIM AJUDEMMMM ​

Answer 1

A combinação $\sf \left( C_{n,m}\right)$ de "n" termos tomados "m" a "m" é dada por:

$\sf C_{n,m}~=~\dfrac{n!}{m!\cdot (n-m)!}$

Então:

$\sf C_{11,4}~=~\dfrac{11!}{4!\cdot (11-4)!}\\\\\\\sf C_{11,4}~=~\dfrac{11!}{4!\cdot 7!}\\\\\\\sf C_{11,4}~=~\dfrac{11\cdot 10\cdot 9\cdot 8\cdot 7!}{4\cdot 3\cdot 2\cdot 1\cdot 7!}\\\\\\\sf C_{11,4}~=~\dfrac{11\cdot 10\cdot 9\cdot 8\cdot \not\!\!7!}{4\cdot 3\cdot 2\cdot 1\cdot \not\!\!7!}\\\\\\\sf C_{11,4}~=~\dfrac{11\cdot 10\cdot 9\cdot 8}{4\cdot 3\cdot 2\cdot 1}\\\\\\\sf C_{11,4}~=~\dfrac{7920}{24}\\\\\\\boxed{\sf C_{11,4}~=~330}$

$\sf C_{7,3}~=~\dfrac{7!}{3!\cdot (7-3)!}\\\\\\\sf C_{7,3}~=~\dfrac{7!}{3!\cdot 4!}\\\\\\\sf C_{7,3}~=~\dfrac{7\cdot 6\cdot 5\cdot 4!}{3\cdot 2\cdot 1\cdot 4!}\\\\\\\sf C_{7,3}~=~\dfrac{7\cdot 6\cdot 5\cdot \not\!\!4!}{3\cdot 2\cdot 1\cdot \not\!\!4!}\\\\\\\sf C_{7,3}~=~\dfrac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}\\\\\\\sf C_{7,3}~=~\dfrac{210}{6}\\\\\\\boxed{\sf C_{7,3}~=~35}$

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