Question

How many 3-digit even numbers can you form such that if one of the digits is $5$ then the following digit must be $7$?

• A. 5
• B. 405
• C. 365
• D. 495

Hint:

When a restriction is given, count total cases and subtract violating cases, adjusting for overlaps using inclusion-exclusion.

Answer 1

The Correct Option is C

Step 1: Total 3-digit even numbers without restriction. The unit's digit can be $0,2,4,6,8$ (5 choices). The hundreds digit can be $1$ to $9$ (9 choices). The tens digit can be $0$ to $9$ (10 choices). So: \[ 9 \times 10 \times 5 = 450 \ \text{total even numbers} \] Step 2: Remove cases violating the rule. The rule says: If digit $5$ appears, the next digit must be $7$. Violation means having $5$ followed by a digit other than $7$. Count violations: - Case 1: Hundreds digit = 5, tens digit $\neq 7$. Tens has $9$ choices, units = even (5 choices): $1 \times 9 \times 5 = 45$. - Case 2: Tens digit = 5, units digit $\neq 7$. But unit is even, so cannot be $7$, hence every such case is a violation. Hundreds = 9 choices, tens = 1, units = 5 choices: $9 \times 1 \times 5 = 45$. Step 3: Overlap case Hundreds = 5, tens = 5, units = even — this violates twice but counted twice, adjust: $1 \times 1 \times 5 = 5$. So total violations = $45 + 45 - 5 = 85$. Step 4: Subtract from total \[ 450 - 85 = 365 \] \fbox{Final Answer: 365} %Quick tip