How many 3-digit numbers can be formed from the digits 1, 2, 3, 5, 6, 7 and 0 which are divisible by

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12 Apr 2017 Permutation

How many 3-digit numbers can be formed from the digits 1, 2, 3, 5, 6, 7 and 0 which are divisible by 5 and none of the digits is repeated?

OPtion

1) 20
2) 24
3) 30
4) 36
5) 40
6) 44
7) 48
8) 55
9) 54
10) None of these

Solution

Answers: 55
To be divisible by 5 unit digit of number should be either 5 or 0
We have total 7 digits out of which we cannot have 0 at the hundred’s place (else it will become a 2 digit number)
Unit place is filled with 5
a. Tens place can be filled with 0
Hundreds place can now be filled with (1, 2, 3, 6, 7). We have 5 ways
So we have 1*1*5 = 5
b. Tens place can be filled with any of the digits (1, 2, 3, 6, 7). We have 5 ways
Hundreds place can now be filled with remaining of 4 digits. We have 4 ways
So we have 1*5*4 = 20

Unit place is filled with 0
Tens place can be filled with any of the digits (1, 2, 3, 5, 6, 7). We have 6 ways
Hundreds place can now be filled with remaining of 5 digits. We have 5 ways
So we have 1*6*5 = 30

So Total number of ways = 5 + 20 + 30 = 55

Option 8)

Correct Option: 8 Best Solution (2)

vinay chouhan   7 years AGO

number is divisible by 5 when unit digit is either 0 or 5
case1) when unit digit is 0
then no. of 3 digit no.=6*5*1=30
case2) when unit digit is 5
then no. of 3 digit no.=5*5*1=25
total 3 digit no.'s possible=25+30=55 ans.

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ganesh chemanchula   7 years AGO

A number to be divisible by 5 should have 0 or 5 in the unit's place.
1. If we keep 0 in units place..then we have 6 chances to fill hundred's place and 5 chances to fill ten's place..as no digit to be repeated..so total numbers will be..6*5*1=30.
2. If we keep 5 in unit's place..then we have 5 chances to fill hundred's place..as 0 can't be in hundred's place and for ten's place we have 5 chances..so total will be 5*5*1=25.
So total numbers will be 30+25=55. So option 8.

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