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

Similar CA Foundation Doubts

The total number of numbers less than 1000 and divisible by 5 formed with 0,1,2,3,.9 such that each digit does not occur more than once in each number is (

Compute the sum of 4 digits number which can be formed from the numbers 1,3,5,7.If each digit is used only once in the arrangement?

4 digit numbers to be formed out of the figures 0, 1, 2, 3, 4 (no digit is repeate d) then number of such numbers is?

Four digits 1,2,4 and 6 are selected at random to form a four digit number. What Is the probability that the number so formed , would be divisible by 4?

The number of 4 digit numbers greater than 5000 can be formed out of the digits 3,4,5,6 and 7 (no. Digit is repeated ). The number of such is?

Top Courses for CA FoundationView all

Quantitative Aptitude for CA Foundation

Principles and Practice of Accounting

Business Laws for CA Foundation

Business Economics for CA Foundation

Mock Tests & Past Year Papers for CA Foundation

Top Courses for CA Foundation

Top Courses for CA Foundation

Suggested Free Tests

Related Content

CA Foundation Courses

Other Courses

How To Prepare?

Other Exams

Company