DemitiAfrorii DemitiAfrorii

09-01-2017
Mathematics

Answered

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?

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LammettHash LammettHash

09-01-2017

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

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

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