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MathematicsPermutation CombinationArrangement under ConstraintMedium2 minPYQ_2021
MathematicsMediumnumerical

The total number of numbers, lying between100and1000that can be formed with the digits1,2,3,4,5,if the repetition of digits is not allowed and numbers are divisible by either3or5,is

Question diagram: The total number of numbers, lying between 100 and 1000 that
Answer:
32.00
Solution:

We need three digits numbers.

Since 1+2+3+4+5=15

So, number of possible triplets for multiple of 15 is 1×2×2

so =4×3+4×3-1×2×2=32

Alternate:

 

divisible by 5=4×3=12

Divisible by 3

123, 4, 53!=6152, 3, 43!=6241, 3, 53!=6421, 2, 33!=6                              24

Divisible by 15

1,4,5; 4,1,5; 3,4,5; 4,3,5

Required No. =24+12-4=32

Stream:JEESubject:MathematicsTopic:Permutation CombinationSubtopic:Arrangement under Constraint
2mℹ️ Source: PYQ_2021

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