Author 
Message 
Intern
Joined: 24 Sep 2008
Posts: 17

How many fivedigit numbers can be formed using the digits [#permalink]
Show Tags
11 Oct 2008, 15:09
How many fivedigit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? * 15 * 96 * 120 * 181 * 216 Can someone explain solution approach for this question. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



Manager
Joined: 09 Oct 2008
Posts: 93

Re: m4  32  [#permalink]
Show Tags
11 Oct 2008, 21:08
IMO E? lets take first digit as 5.. so remaining 4 digit can be arrange in this pattern pattern1: 5 0 4 2 1 ...sum =12 /3 ..okay..so total possible combination 1*4!= 24 pattern1: 5 4 3 2 1 ...sum =15 /3 ..okay..so total possible combination 1*4!= 24 so for if first digit is 5 we have 48 numbers lets take first digit as 4.. so remaining 4 digit can be arrange in this pattern pattern1: 4 0 5 2 1 ...sum =12 /3 ..okay..so total possible combination 1*4!= 24 pattern1: 4 5 3 2 1 ...sum =15 /3 ..okay..so total possible combination 1*4!= 24 so for if first digit is 4 we have 48 numbers same goes on for all the number and you willend up with 48*5 =240..which is not in answer so bestnear by answer is 216 * i will follow this proc.when i have very less time say 90sec. in exam and i have to mark some answer so i will go for E. i knowi madesomesmallmistake somewhere..please rectify it



SVP
Joined: 29 Aug 2007
Posts: 2452

Re: m4  32  [#permalink]
Show Tags
11 Oct 2008, 21:48
2
This post received KUDOS
lalmanistl wrote: How many fivedigit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
* 15 * 96 * 120 * 181 * 216
Can someone explain solution approach for this question. 5 digit integers made of "1, 2, 3, 4, and 5" & "0, 1, 2, 4, and 5" are divisible by 3. so the no. of possibilities are = 5! + 4x4! = 120 + 96 = 216
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Manager
Joined: 10 Aug 2008
Posts: 73

Re: m4  32  [#permalink]
Show Tags
12 Oct 2008, 04:13
5 digit number formed which is divisible by 3 can be formed by combonation of
1,2,3,4&5 in 5! ways OR 0,1,2,4 & 5 with 0 will not be Most Significant digit in 4X4!
Total ways = 216



Intern
Joined: 12 Oct 2008
Posts: 1

Re: m4  32  [#permalink]
Show Tags
12 Oct 2008, 07:27
Only when we are using the set of numbers 0,1,2,4 & 5 or 1,2,3,4 & 5 should the 5 digit number be divisible by 3
with the first set if we remove the 5 digits set that will start with 0 that is 24 from 5! for the whole probability then the difference will be 5!  4! = 120  24 = 96
then we add the number of probabilities for the second set
therefore we get the result of
5! + 5!  4! = 120 + 120 24 = 240  24 = 216
thanks
Mohamed



Manager
Joined: 02 Aug 2007
Posts: 221
Schools: Life

Re: m4  32  [#permalink]
Show Tags
27 Jan 2009, 20:55
GMAT TIGER wrote: lalmanistl wrote: How many fivedigit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
* 15 * 96 * 120 * 181 * 216
Can someone explain solution approach for this question. 5 digit integers made of "1, 2, 3, 4, and 5" & "0, 1, 2, 4, and 5" are divisible by 3. so the no. of possibilities are = 5! + 4x4! = 120 + 96 = 216 Can someone tell me why we multiply 5*4*3*2*1? and why we multiply 4*3*2*1? is there a rule for using factorial in creating number combinations?



Manager
Joined: 04 Jan 2009
Posts: 234

Re: m4  32  [#permalink]
Show Tags
28 Jan 2009, 10:45
We have 5 ways to fill up the first place:5 5 ways to fill up the second :5 4 ways to fill up the thrid:4 3 ways to fill up the fourth:3 and 2 ways to fill up the fifth. Multiplying, I get 600. What is wrong with this approach? xALIx wrote: GMAT TIGER wrote: lalmanistl wrote: How many fivedigit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
* 15 * 96 * 120 * 181 * 216
Can someone explain solution approach for this question. 5 digit integers made of "1, 2, 3, 4, and 5" & "0, 1, 2, 4, and 5" are divisible by 3. so the no. of possibilities are = 5! + 4x4! = 120 + 96 = 216 Can someone tell me why we multiply 5*4*3*2*1? and why we multiply 4*3*2*1? is there a rule for using factorial in creating number combinations? == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
_________________
 tusharvk










