A= (0,1,2,3,4,5). how many 3 digit numbers can be made using digits of set A such that number is divisible by 5 ?
A. 40 B. 60 C. 120 D. 180 E. 216
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Permutation Problem
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kvcpk
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A= (0,1,2,3,4,5)akpareek wrote:A= (0,1,2,3,4,5). how many 3 digit numbers can be made using digits of set A such that number is divisible by 5 ?
A. 40 B. 60 C. 120 D. 180 E. 216
First position can be filled by 1,2,3,4,5 - 5 ways
Second position can be filled by 0,1,2,3,4,5 - 6 ways
Third position can be filled by 0,5 - 2 ways
Hence total = 5*6*2 = 60 ways
Hope this helps!!
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clock60
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here i got 60
it can be done two ways
1. total number of 3 digit numbers with digits 0,1,2,3,4,5 ( assuming no restrictions) is 5*6*6=180
among then 5*6*4=120 do not end on 5, or 0.
180-120=60
another way
numbers ending with 5, 5*6*1=30
numbers ending with 0, 5*6*1=30
30+30=60
it can be done two ways
1. total number of 3 digit numbers with digits 0,1,2,3,4,5 ( assuming no restrictions) is 5*6*6=180
among then 5*6*4=120 do not end on 5, or 0.
180-120=60
another way
numbers ending with 5, 5*6*1=30
numbers ending with 0, 5*6*1=30
30+30=60
thank you very much.......... i understood how to solve it....i think it is shortcut method .... your simply great........kvcpk wrote:A= (0,1,2,3,4,5)akpareek wrote:A= (0,1,2,3,4,5). how many 3 digit numbers can be made using digits of set A such that number is divisible by 5 ?
A. 40 B. 60 C. 120 D. 180 E. 216
First position can be filled by 1,2,3,4,5 - 5 ways
Second position can be filled by 0,1,2,3,4,5 - 6 ways
Third position can be filled by 0,5 - 2 ways
Hence total = 5*6*2 = 60 ways
Hope this helps!!
