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12. Suppose that a combination lock is opened by entering a three-digit code. Each digit can be any integer between 0 and 9, but digits may not be repeated in the code. How many
different codes are possible? Is this question answered by considering permutations or combinations? Explain.

Respuesta :

Answer:120

Step-by-step explanation:

from 0 - 9, we have 10 alphabets.

nCr = n!/ (n-r)!r!

10C3 = 10!/ (10-3!)3!

=10 × 9×8×7! /7!3!

=10×9×8 /6

=720/6

=120

The answe to the second part of the question:

The question is answered considering combination because in the arrangement of the digit, order doesn't matter in the arrangement.

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