macarsovoAngiekola macarsovoAngiekola

12-12-2016
Mathematics

contestada

Suppose that a "code" consists of 4 digits, none of which are repeated. (a digit is one of the 10 numbers 0,1,2,3,4,5,6,7,8,9) how many codes are possible?

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JcAlmighty JcAlmighty

15-12-2016

The answer is 5040.

This is permutation without the repetition:
P = n!/(n - r)!

n - the number of choices
r - the number of chosen

We have:
n = 10
r = 4

P = 10!/(10 - 4)!
= 10!/6!
= (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (6 * 5 * 4 * 3 * 2 * 1)
= 10 * 9 * 8 * 7
= 5040

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