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  • Usuário do Brainly
2014-09-04T02:35:31-03:00
A_{n,k}=\dfrac{n!}{(n-k)!}

a) A_{x,2}=90

\dfrac{x!}{(x-2)!}=90

Veja que, x!=x(x-1)(x-2)!. Assim:

\dfrac{x(x-1)(x-2)!}{(x-2)!}=90

x(x-1)=90

Mas, 90=10\cdot9, donde, x=10


b) A_{x,3}=A_{x,2}

\dfrac{x!}{(x-3)!}=5\cdot\dfrac{x!}{(x-2)!}

Note que, x!=x(x-1)(x-2)(x-3)!, então:

\dfrac{x(x-1)(x-2)(x-3)!}{(x-3)!}=5\cdot\dfrac{x(x-1)(x-2)!}{(x-2)!}

x(x-1)(x-2)=5\cdot x(x-1)

\dfrac{x(x-1)(x-2)}{x(x-1)}=\dfrac{5\cdot x(x-1)}{x(x-1)}

x-2=5

x=7
5 4 5