1. **Problem statement:** We have digits 2, 4, 5, 7, and 9 and want to find: - Total numerals formed using all digits without repetition. - How many are multiples of 5. - How many are even. - How many are greater than 70,000. - How many are less than 50,000. 2. **Total numerals formed:** Using all 5 digits without repetition means arranging 5 distinct digits in all possible orders. Formula: Number of permutations of $n$ distinct digits = $n!$ Here, $n=5$, so total numerals = $5! = 120$ 3. **Multiples of 5:** A number is divisible by 5 if its last digit is 0 or 5. Since 0 is not in our digits, last digit must be 5. Fix last digit as 5, arrange remaining 4 digits in any order. Number of such numbers = $4! = 24$ 4. **Even numbers:** Even numbers end with an even digit. Our even digits are 2 and 4. Case 1: Last digit is 2, arrange remaining 4 digits in any order: $4! = 24$ Case 2: Last digit is 4, arrange remaining 4 digits: $4! = 24$ Total even numbers = $24 + 24 = 48$ 5. **Numbers greater than 70,000:** The first digit must be 7 or 9 (since 2,4,5 are less than 7). Case 1: First digit 7, arrange remaining 4 digits: $4! = 24$ Case 2: First digit 9, arrange remaining 4 digits: $4! = 24$ Total numbers > 70,000 = $24 + 24 = 48$ 6. **Numbers less than 50,000:** The first digit must be 2 or 4 (digits less than 5). Case 1: First digit 2, arrange remaining 4 digits: $4! = 24$ Case 2: First digit 4, arrange remaining 4 digits: $4! = 24$ Total numbers < 50,000 = $24 + 24 = 48$ **Final answers:** - Total numerals: 120 - Multiples of 5: 24 - Even numbers: 48 - Numbers > 70,000: 48 - Numbers < 50,000: 48