Question

A key code must contain 6 numerals. There are 10 numerals available. Using these numerals, how many different key codes may be created? A. 151,200 B. 340 C. 210 D. 3,480

Answer 1

$10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5=151,200$

Answer 2

Answer: Option A, 151,200 possible combinations.

Explanation:

The code must contain 6 numbers, and there are a total of 10 numbers avaible. Here we must assume that the numbers must be different, because of the word "avaible" and the options avaible.

This means that for the first number, we have 10 options, for the second number, we have 9 options (because we already took 1), in the third number, we have 8 options, and so on.

The total number of combinations is equal to the product between the number of options for each number in the code.

10*9*8*7*6*5 = 151,200

Then the right option is A.