Question

The number of five-digit numbers formed using the digits \(2,3,5,7,9\) without repetition and which are greater than \(24000\) are

• A. \(120\)
• B. \(117\)
• C. \(114\)
• D. \(96\)

Hint:

Count all 120 and subtract numbers starting with 23.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We count five-digit numbers from the digits 2, 3, 5, 7, 9 (each used once) that exceed 24000. There are \(5! = 120\) numbers in all.

Step 2: Case 1: first digit is 3, 5, 7 or 9:
Any of these four digits first makes the number at least 30000, which is above 24000. The remaining 4 digits can be arranged in \(4! = 24\) ways. This gives \(4\times 24 = 96\) numbers.

Step 3: Case 2: first digit is 2:
The other digits are 3, 5, 7, 9. If the second digit is 3, the number is \(23xxx\), which is less than 24000. This happens in \(3! = 6\) ways. If the second digit is 5, 7 or 9, the number is at least \(25xxx\), which is above 24000. This gives \(3\times 3! = 18\) numbers.

Step 4: Total:
\[ 96 + 18 = 114 \]

Step 5: Choose:
Option (C). The value 120 counts every arrangement without applying the condition, and 117 would result from excluding only 3 numbers instead of 6.

Final Answer:
There are 114 such numbers, option (C). \[ \boxed{114} \]