The number of five-digit numbers, greater than 40000 and divis ible by 5, which can be formed using the digits 0, 1, 3, 5, 7, and 9 without repetition, is equal to:
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When dealing with permutations and combinations with restrictions (like divis ibility and a specific range), break the problem into cases based on the restric tions. This makes the problem easier to solve.
A five-digit number is divisible by 5 if its last digit is either 0 or 5. The number must also be greater than 40000.
Case 1: The last digit is 0 - If the last digit is 0, the first digit can be 5, 7, or 9 (since the number must be greater than 40000). This gives us 3 choices for the first digit. - For the remaining three digits, we have 4 remaining choices (we’ve used two digits already), and these can be arranged in: \[ 4 \times 3 \times 2 = 4P3 = 24 \, \text{ways}. \] - So, the number of five-digit numbers ending in 0 is: \[ 3 \times 24 = 72. \]
Case 2: The last digit is 5 - If the last digit is 5, the first digit can be 7 or 9 (since the number must be greater than 40000, and we can’t use 5 again). This gives us 2 choices for the first digit. - For the remaining three digits, we have 4 remaining choices, and they can be arranged in: \[ 4 \times 3 \times 2 = 4P3 = 24 \, \text{ways}. \] - So, the number of five-digit numbers ending in 5 is: \[ 2 \times 24 = 48. \]
Total Number of Five-Digit Numbers: Adding the counts from both cases, we get: \[ 72 + 48 = 120. \]
Final Answer: There are 120 such five-digit numbers.
