digits

[ketkoag]

Guys, this question has already been asked by hk but from the solutions there i am not able to get it.. please help..

How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

a) 15
b) 96
c) 120
d) 181
e) 216

[rahulg83]

My answer 216

consider no as --,--,--,--,-- or in other words ABCDE

now in this number, either you can use 3 or 0, but not both. Because if you use both of them, you have to remove one number in 1,2,4,5

ex 12345 is divisible by 3
12045 also is divisible by 3
but 12305?? nopes (Logic 1+2+3+4+5+0 = 15, divisible by 3, remove either 0 or 3 to keep it divisible by 3)

So consider case 1, suppose you put A=3, only 1 option to do this,
now remember we can't use 0 in this number, so B has four options (1,2,4,5), C has 3, D has 2 and E has 1. therefore total no of cases = 1*4*3*2*1=24

case 2, Let us assume we are not using 0 in our numberA has either (1,2,4 or 5), 4 options to do this. , B has 4 options (we can use 3 and three other digits) , C has 3, D has 2, E has 1
total no of cases = 4*4*3*2*1 = 96

case 3 Let us assume we are not using 3 in our number, no of combination will be exactly as in case 2, 96

Therefor total cases = 96+96+24 =216

[ketkoag]

my approach is :
we can take either 3 or 0 at a time.. so that the sum of all the digits is divisible by 3 and hence the no. will be divisible by 3.
so case 1 : we take 3.

_ _ _ _ _ or 5 * 4 *3 *2 *1 these are the possibilities to make a 5 digit number with letters 1,2,3,4,5 = 120

case 2: we take 0.
_ _ _ _ _ or 4 * 4 *3 *2 *1 these are the possibilities to make a 5 digit number with letters 1,2,0,4,5 = 96

so 120 + 96 = 216 ways.

got it when tried it today..
nyway thanks..