JEE Main 2022MathematicsPermutation CombinationHardNumerical

JEE Main 2022Permutation Combination Question with Solution

JEE Main 2022 (29 Jul Shift 2)

Question

The number of natural numbers lying between 1012 and 23421 that can be formed using the digits 2,3,4,5,6 (repetition of digits is not allowed) and divisible by 55 is _____.

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Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation

Given number should be 4 digit numbers between 1012 & 23421 and can be formed using the number 2, 3, 4, 5, & 6 and should be divisible by 55

CASE I: When number has 4 digit, 

Now for divisibility by 55, number should be divisible by 5 and 11 both

Also, for divisibility by 11 for below figure,

Now fixing d=5 the total possible cases for a,b,c will be 6,4,3,3,4,6,2,3,6,6,3,2,3,2,4 & 4,2,3

CASE II:

When number has 5 digits.

Let the number be abcde.

Here e is fixed at 5. So the other 4 digits must be from 2,3,4,6

No such number is possible because even the least number formed is greater than 23421.

Hence, the total number of required numbers =6

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About this question

This is a previous-year question from JEE Main 2022, covering the Permutation Combination chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.