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Combinations

by hk » Fri Jun 12, 2009 7:43 pm
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
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by ssmiles08 » Fri Jun 12, 2009 8:04 pm
[Edited] b/c the way I approached the entire thing was wrong. [Edited]
Last edited by ssmiles08 on Sat Jun 13, 2009 5:17 am, edited 1 time in total.
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Re: Combinations

by Naruto » Fri Jun 12, 2009 8:07 pm
hk wrote: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
IMO E

Ok, so for a no. to be divisible by 3, sum of the digits should be divisible by 3. That divides the given nos in two groups to form the 5 digit no. required.
i.e Group A comprising of 1,2,3,4,5
and
Group B comprising of 0.1.2.4.5

Now for group A the possiblity without repeating a digit for a 5 digit no. is
5X4X3X2X1= 120
Similarly for group B,(Note , since this group has a 0, the first digit can't be zero, hence)
4X4X3X2X1= 96

Now our answer is the sum of two, i.e 120+96= 216.

Let me know the OA.
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by gmatplayer » Sat Jun 13, 2009 4:03 am
I think it is
= (total number of ways all digits can be placed anywhere) - (total number of ways zero cannot be placed as first digit)

= 5!-4!

=120-24

=96

There is No need to find out where formed number is divisible by 3 because the sum is always divisible by 3.

Try this just by taking digit 0, 1 and 2
= 3!-2!

=4
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by tohellandback » Sat Jun 13, 2009 5:12 am
IMO 216

the set of numbers can be-0,1,2,4,5 or 1,2,3,4,5
number of 5 digit numbers with 1,2,3,4,5=5!=120

number of 5 digit numbers with 0,1,2,4,5= 5!-4!(numbers starting with 0)
= 96
120+96=216
The powers of two are bloody impolite!!
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by gmatplayer » Sat Jun 13, 2009 5:40 am
You are right

There are six digits, silly me
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