The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, and 5 is 54.
To determine the total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, and 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, we can break down the problem into two cases: numbers divisible by 3 and numbers divisible by 5.
Case 1: Numbers divisible by 3
For a number to be divisible by 3, the sum of its digits must be divisible by 3. In this case, we have five digits to choose from: 1, 2, 3, 4, and 5. Using these digits, we can form three-digit numbers that are divisible by 3 in the following ways:
Choose three different digits from the five available options: 5C3 = 10 ways
Choose two of the same digit and one different digit: 5C2 * 4 = 40 ways
Therefore, the total number of three-digit numbers divisible by 3 that can be formed with the digits 1, 2, 3, 4, and 5, without repetition, is 10 + 40 = 50.
Case 2: Numbers divisible by 5
For a number to be divisible by 5, the last digit must be either 0 or 5. Since we are not allowed to repeat digits, we have only one option for the last digit: 5. This means that the first two digits can be chosen from the remaining four digits: 4C2 = 6 ways.
Therefore, the total number of three-digit numbers divisible by 5 that can be formed with the digits 1, 2, 3, 4, and 5, without repetition, is 6.
Total Number of Numbers
To find the total number of numbers that satisfy both conditions (divisible by 3 or 5), we need to subtract the numbers that are divisible by both 3 and 5. There are two such numbers: 150 and 300.
Therefore, the total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, and 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is 50 + 6 - 2 = 54.