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A 5-digit security code has the numbers 1 through  7 to choose from. If each number can only be used once, how many different combinations are possible?

Respuesta :

LammettHash
The first position has 7 options to choose from, the second has 6, the third has 5, and so on. So the number of different arrangements is

[tex]_{7}P_{5}=5!\dbinom75=5!\dfrac{7!}{5!(7-5)!}=\dfrac{7!}{2!}=2520[/tex]

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