A briefcase has a three-digit lock code that does not include zero as a digit.
What is the probability that the lock code consists of all even digits if the same digit is not used more than once in the lock
code?
24 out of 504
O 60 out of 504
24 out of 84
O 60 out of 84

Id say 24 out of 504

Because there would be a whole lot of combinations so 24 out of 504 sounds reasonable.

Choice A

24 out of 504

This is the same as writing the fraction 24/504

Work Shown:

set of even numbers allowed = {2,4,6,8}

we do not include 0 as that is stated in the instructions

For slot A, we have 4 even numbers to pick from. Then for slot B, there are 4-1 = 3 even numbers. Then slot C has 2 even numbers.

We have 4*3*2 = 12*2 = 24 different permutations possible of picking all even digits and there are no repeats.

This is out of 9*8*7 = 504 different codes where we do not repeat a digit (odd digits are allowed now).

So this is why the answer is 24/504 or "24 out of 504".

99

step-by-step explanation:

step-by-step explanation:

0