Question

A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not consist of all odd digits if the same digit is not used more than once in the code?

Answer 1

The answer is D: 4,920 out of 5,040

Answer 2

Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.