Respuesta :

WojtekR

We know that:

[tex]_nP_k = \dfrac{n!}{(n-k)!}[/tex]

so in this case [tex]n = 5[/tex], [tex]k=3[/tex] and:

[tex]_5P_3=\dfrac{5!}{(5-3)!}=\dfrac{5!}{2!}=\dfrac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\dfrac{120}{2}=\boxed{60}[/tex]

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