Ashie Posted January 21, 2005 Share Posted January 21, 2005 How many number of 3 digit numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number? How many of these are even and how many odd? Please explain. Quote Link to comment Share on other sites More sharing options...
shanmukh Posted January 21, 2005 Share Posted January 21, 2005 60 odd and 60 even Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted January 21, 2005 Share Posted January 21, 2005 5*5*4 = 100 in Total (1st in 5 ways, leaving 0, left are 5 for 2nd digit & 4 for last) __ __ __ Odd: fix last as odd, 3 ways __ __ _3_ now, left are 5, but again leaving 0, 4 for 1st digit & again 4 foir 2nd digit: _4_ _4_ _3_ =48 Odd. 100-48= 52 Even Quote Link to comment Share on other sites More sharing options...
Ashie Posted January 24, 2005 Author Share Posted January 24, 2005 Thanks Bullfighter. Yep, the answer is what you have given. A small clarification - is it that we have to always find the number of Odd first, then do 100-48 to get the even? If we have to calculate even first, how do we do that? I know it's easier to find odd and then subtract from 100 to get even. But still would like to know how to go about the other way. Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted January 24, 2005 Share Posted January 24, 2005 you can find either way, But, finding # even is not as simple as finding # odd. Here, we have to consider Two diff. cases. One when 0 is used at the last digit, & hence automatically can not be used at the 1st place. & the other one where 0 is Not used at the last place... Even: For last digit: 0, 2, 4 1. Last digit: 2 & 4. Other left: 0,1,3,5. 4*4*2= 32 2. Last as 0. Other left: 1,2,3,4,5. 5*4*1= 20 Total= 52. Odd= 100-52= 48. Quote Link to comment Share on other sites More sharing options...
crazyonMBA Posted January 24, 2005 Share Posted January 24, 2005 Bull, Can't figure out and how u solved the question? Can you pls xplain a bit .. Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted January 24, 2005 Share Posted January 24, 2005 for the original Q, How many number of 3 digit numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number? Choices available:0,1,2,3,4,5. ie. Six in total. 1000'th digit can be filled in 5 ways (as 0 not allowed), now, repetition not allowed, hence left ar 5 digits, Hence, 100th digit can also be filled in 5 ways. Again, left are 4, hence Unit's digit can be filled in 4 ways. Hence, total ways: 5*5*4 (Permutation) ___________________________ for Odd. Unit's digit can be filled in 3 ways. (1,3,5 As Odd) Left are 5 digits, again as 0 can not be used, for 1000th digit choices left are 4. (one used as Unit digit & also excluding 0, 6-2=4) & for 100th digit, 2 already used, one at unit's place & one at 1000th place, laft are 4 choices. Hence, 4*4*3 Similarly for Even, as shown above. Quote Link to comment Share on other sites More sharing options...
crazyonMBA Posted January 24, 2005 Share Posted January 24, 2005 Wow, Gr8 Quote Link to comment Share on other sites More sharing options...
Ashie Posted January 24, 2005 Author Share Posted January 24, 2005 you can find either way, But, finding # even is not as simple as finding # odd. Here, we have to consider Two diff. cases. One when 0 is used at the last digit, & hence automatically can not be used at the 1st place. & the other one where 0 is Not used at the last place... Even: For last digit: 0, 2, 4 1. Last digit: 2 & 4. Other left: 0,1,3,5. 4*4*2= 32 2. Last as 0. Other left: 1,2,3,4,5. 5*4*1= 20 Total= 52. Odd= 100-52= 48. Thanks a bunch Bullfighter! Quote Link to comment Share on other sites More sharing options...
n_adroja Posted January 24, 2005 Share Posted January 24, 2005 Hi bullfighter, Thanks for ur explanation. However, Can u pls tell me why did u split the method of finding EVEN numbers into two parts. One where 0 is at the end & the other where it is not? Y can we not consider it as one option - If it has to be even then the last digit should be any of the three 0,2,4. so there are 3 possible ways for third digit, 4 for first 4 for second what is wrong in this method? I know the ans does not match ,it is just that I want to figure out what is missed in this approach Can u pls help me out thanks adroja you can find either way, But, finding # even is not as simple as finding # odd. Here, we have to consider Two diff. cases. One when 0 is used at the last digit, & hence automatically can not be used at the 1st place. & the other one where 0 is Not used at the last place... Even: For last digit: 0, 2, 4 1. Last digit: 2 & 4. Other left: 0,1,3,5. 4*4*2= 32 2. Last as 0. Other left: 1,2,3,4,5. 5*4*1= 20 Total= 52. Odd= 100-52= 48. Quote Link to comment Share on other sites More sharing options...
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