A personal identification number (PIN) that opens a certain lock consists of a sequence of 3 different digits from 0 through 9, inclusive. How many possible PIN are there? (A) 120 (B) 360 (C) 720 (D) 729 (E) 1,000

720 possible PIN can be generated.

Explanation:

To calculate different number of orders of digits to create password and PIN, we calculate permutation.

Permutation is a term that means the number of methods or ways in which different numbers, alphabets, characters and objects can arranged or organized. To calculate the permutation following formula will be used:

nPr = n! / (n-r) !

there P is permutation, n is number of digits that need to be organize, r is the size of subset (Number of digits a password contains)

So in question we need to calculate

P=?

where

n = 10 (0-9 means total 10 digits)

r = 3 (PIN Consist of three digits)

So by using formula

10P3 = 10! / (10-3) !

=10!/7!

= 10x9x8x7!/7!

= 10x9x8

= 720