Mathematics - Permutation & Combination Question with Solution | TestHub

MathematicsPermutation & CombinationnPr & nCrMedium2 minPYQ_2022
MathematicsMediumnumerical

Numbers are to be formed between1000and3000, which are divisible by4, using the digits1,2,3,4,5and6without repetition of digits. Then the total number of such numbers is _______.

Answer:
30.00
Solution:

To find total numbers between 1000 and 3000 divisible by 4 using the digits 1,2,3,4,5,6,

We will solve in two cases.

Case I : When first digit is 1.

Then last two digits can be 24,32,36,52,56 or 64

So, total ways of choosing last two digit is 6 and second digit will be chosen in 3 ways

So, number of such numbers =6×3=18

Case II: When first digit is 2

Then last two digits can be 16,36,56 or 64

So, total ways of choosing last two digit is 4 and second digit will be chosen in 3 

So, number of such numbers =4×3=12

Total numbers of numbers =18+12=30

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:nPr & nCr
2mℹ️ Source: PYQ_2022

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