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?

There are 10 choices for the first digit, then 9, 8, and 7 choices for the remaining digits, giving 10 × 9 × 8 × 7 = 5,040 possible codes. A code made entirely of odd digits has 5 × 4 × 3 × 2 = 120 possibilities, since there are five odd digits and digits cannot repeat. Therefore, 5,040 − 120 = 4,920 codes do not consist of all odd digits, so the probability is 4,920 out of 5,040, choice C.

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