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A bank PIN has (5) digits. The first digit cannot be (0), and the PIN must contain at least one digit (7). Digits may repeat. How many PINs are possible?

Answer and explanation

Correct answer: 37512

Use the complement method. There are 9 choices for the first digit (1 to 9) and 10 choices for each of the remaining four positions, so the total number of PINs is \(9 \times 10^4=90000\). For PINs with no 7 at all, the first digit has 8 choices (1 to 9 except 7), while each remaining position has 9 choices. Thus, there are \(8 \times 9^4=52488\) such PINs. Therefore, the number containing at least one 7 is \(90000-52488=37512\). Note that 52488 counts PINs containing no 7. Exam tip: for “at least one” counting questions, try total minus none first.

Related tags

Fundamental Counting PrincipleComplement CountingBank PinDigit ArrangementsPermutations And CombinationsFundamental Principle Of CountingMathematicsClass 11 Mcq

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