Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET

Best NIMCET Coaching
Aspire Study
Online Classes, Classroom Classes
and More.

Previous Year Question (PYQs)

Let n be the number of different 5 digit numbers, divisible by 4 that can be formed with the digits 1,2, 3, 4, 5 and 6, with no digit being repeated. What is the value of n ?




Practice With Us

Practice Makes A Man Perfect. Comment your solution and approach for this question.

Video solution of this question given below

We have to make 5 digit no. And also divisible by 4 using digits (1,2,3,4,5,6) To be divisible by 4 last two digit of 5 digits no should be divisible by 4 so we choose last two digits st they are divisible by 4 and rest with remaining four digits. Last two digits only can be 12,16,24,32,36,52,56,64 as all are divisible by 4. So total no of ways= no ways of choosing first 3 digits* no ways of choosing last two digits Total= (4*3*2)*(8) =192

Abhijeet Kushwaha

Aspire Study Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Aspire Study Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More
Ask Your Question or Put Your Review.

loading...
Abhijeet Kushwaha -pic
Abhijeet Kushwaha , Student
Commented Feb 09 , 2023
We have to make 5 digit no. And also divisible by 4 using digits (1,2,3,4,5,6) To be divisible by 4 last two digit of 5 digits no should be divisible by 4 so we choose last two digits st they are divisible by 4 and rest with remaining four digits. Last two digits only can be 12,16,24,32,36,52,56,64 as all are divisible by 4. So total no of ways= no ways of choosing first 3 digits* no ways of choosing last two digits Total= (4*3*2)*(8) =192
Shivam Gupta-pic
Shivam Gupta , Aspire
Commented Feb 16 , 2023
Thankyou for contributing your solution.

Your reply to this comment :


loading...