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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asked Sep 10, 2018 88.5k views

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Toshiya · Answer 1 · 2018-09-15T13:11:22+0000

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

$_(7)P_(5)=5!\dbinom75=5!(7!)/(5!(7-5)!)=(7!)/(2!)=2520$

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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