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by akpareek » Mon Aug 09, 2010 11:59 am
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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by kvcpk » Mon Aug 09, 2010 12:05 pm
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
A= (0,1,2,3,4,5)
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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by clock60 » Mon Aug 09, 2010 12:10 pm
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
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by aarati » Mon Aug 09, 2010 10:52 pm
kvcpk wrote:
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
A= (0,1,2,3,4,5)
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!!
thank you very much.......... i understood how to solve it....i think it is shortcut method .... your simply great........
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