The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,….9 such that each digit does not occur more than once in each number is?

Question

Answer

Total Numbers Divisible by 5 with Unique Digits

Understanding the problem

The problem requires us to find the total number of numbers less than 1000 and divisible by 5, formed with 0,1,2,….9 such that each digit does not occur more than once in each number.

Approach

We can solve this problem by breaking it down into smaller parts and then combining the results to arrive at the final answer. We can use the following approach:

Count the total number of 3-digit numbers that can be formed using the digits 0-9 without repetition.

Count the total number of 3-digit numbers that are divisible by 5 using the digits 0-9 without repetition.

Combine the results from step 1 and step 2 to arrive at the final answer.

Solution

Step 1: Counting the total number of 3-digit numbers that can be formed using the digits 0-9 without repetition.

To count the total number of 3-digit numbers that can be formed using the digits 0-9 without repetition, we can use the permutation formula:

nP3 = n! / (n - 3)!

Where n is the total number of digits (10) and 3 is the number of digits in each number.

Substituting the values, we get:

10P3 = 10! / (10 - 3)! = 10 x 9 x 8 = 720

So, there are 720 3-digit numbers that can be formed using the digits 0-9 without repetition.

Step 2: Counting the total number of 3-digit numbers that are divisible by 5 using the digits 0-9 without repetition.

To count the total number of 3-digit numbers that are divisible by 5 using the digits 0-9 without repetition, we need to first identify the possible last digits that can make a number divisible by 5. The possible last digits are:

0

5

Since each digit can be used only once, we can use the remaining 8 digits to form the first two digits of the number. So, the total number of 3-digit numbers that are divisible by 5 using the digits 0-9 without repetition is:

2 x 8P2 = 2 x 8! / (8 - 2)! = 2 x 8 x 7 = 112

Step 3: Combining the results from step 1 and step 2 to arrive at the final answer.

From step 1, we know that there are 720 3-digit numbers that can be formed using the digits 0-9 without repetition. From step 2, we know that there are 112 3-digit numbers that are divisible by 5 using the digits 0

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