e.cartman Posted August 28, 2008 Share Posted August 28, 2008 How many different numbers greater than 50,000 can be formed with the digits 1,1,5,9,5,9,0? I don't have the OA. But I tried listing brute force and got 32 of them before growing tired and unsure that there may be more. So I expect answer to be >= 32. Quote Link to comment Share on other sites More sharing options...
e.cartman Posted August 28, 2008 Author Share Posted August 28, 2008 I think I ignored 6 and 7 digit numbers altogether! The number of permutations will be huge. I think this problem is wrong for gre because it will get too hard for the 5-digit case itself, so I won't worry about it. Quote Link to comment Share on other sites More sharing options...
mewidyu Posted August 28, 2008 Share Posted August 28, 2008 yeah may be yu r right.. GRE is not going to ask such lengthy calculations.. i just read the question when yu posted it but wasnt able to solve it... so leaved it.... Quote Link to comment Share on other sites More sharing options...
divya.yadav Posted August 28, 2008 Share Posted August 28, 2008 Its lengthy calculation. Quote Link to comment Share on other sites More sharing options...
e.cartman Posted August 28, 2008 Author Share Posted August 28, 2008 upper limit exists. the number of digits here a finite list, you can only use upto 2 1's, 2 5's, and so on.. Quote Link to comment Share on other sites More sharing options...
vvv134 Posted August 28, 2008 Share Posted August 28, 2008 i have a problem understanding the question.does it mean that all the given digits must used in forming the number.in that case,any number formed will be >50000.so the answer will be 7p4.also it doesnt say each digit can only be used once,in which case the problem gets more complicated. Quote Link to comment Share on other sites More sharing options...
chestnut.cc Posted August 28, 2008 Share Posted August 28, 2008 How many different numbers greater than 50,000 can be formed with the digits 1,1,5,9,5,9,0? I don't have the OA. But I tried listing brute force and got 32 of them before growing tired and unsure that there may be more. So I expect answer to be >= 32Keeping one 5 fixed in the first place we can arrange the remaining 6 nos in 6P5 ways... ie 720 diff nos exist... Quote Link to comment Share on other sites More sharing options...
e.cartman Posted August 28, 2008 Author Share Posted August 28, 2008 Keeping one 5 fixed in the first place we can arrange the remaining 6 nos in 6P5 ways... ie 720 diff nos exist... No, you are allowed to use both 1s, both 5s and so on. But 720 is too big for a 6 digit number because you are not accounting for the fact that there are two 1s and two 9s....You are double-counting by interchanging position of two 9s among themselves,two 1s among themselves, and so on and getting too high an answer. Moreover, the number of digits can vary from 5 to 7 for being more than 50000, not just 6-digits. Quote Link to comment Share on other sites More sharing options...
e.cartman Posted August 28, 2008 Author Share Posted August 28, 2008 i have a problem understanding the question.does it mean that all the given digits must used in forming the number.in that case,any number formed will be >50000.so the answer will be 7p4.also it doesnt say each digit can only be used once,in which case the problem gets more complicated. Yeah, I have seen two flavors of these problems. One where they specify the digits and you can use as many of those digits as you want in fixed number of places. The other is where they give you a list of digits, you pick a digit, you lose that. Since they gave two 9s, you can pick one 9, two 9s or no 9s at all in the number you form. It does not mean all digits must be used. But any of the given digits may be used in given ration. Number of digits in number formed can vary from 0 to 7. Quote Link to comment Share on other sites More sharing options...
e.cartman Posted August 28, 2008 Author Share Posted August 28, 2008 If you think this problem is too hard since there are too many sub-problems you have to divide it into, then imo, you may be on the right track :-) Quote Link to comment Share on other sites More sharing options...
chestnut.cc Posted August 28, 2008 Share Posted August 28, 2008 hmmm... we could then count the no of nos> 50000 which have 5 digits along with all possible nos with 6 digits and then all poss nos with 7 digits... :s Quote Link to comment Share on other sites More sharing options...
e.cartman Posted August 28, 2008 Author Share Posted August 28, 2008 Actually it gets ugly, but let me post my general approach. It may be far from optimal but I don't know any better. 5 digit numbers: Combinations starting with '5' and leaving other 5: ‘5’ + other 4 digits different, sub-permutate ‘5’+ 2 different 2 same, sub-permutate ‘5’ + 2 same 2 same, sub-permutate Two '5'+'5' digit combinations: ‘5’+’5’ + 3 different, sub-permutate ‘5’+’5’+2 same, and 1 different, sub-permutate Then double all above cases for starting with 9. Then solve 6-digit case without 0 in first place 7 digit case without 0 in first place Grand total all permutes. Then hope to never see this problem again in life. Quote Link to comment Share on other sites More sharing options...
rakibeeebuet Posted September 1, 2008 Share Posted September 1, 2008 i think the problem is not relevant for GRE. it must take few minutes to solve. Quote Link to comment Share on other sites More sharing options...
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