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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Mar 2008 19:28:54 IST
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How many numbers between 1 and 1000000 have sum of digits equal to 18?
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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11 Mar 2008 19:34:34 IST
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edited.
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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11 Mar 2008 19:36:03 IST
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25927 ???
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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11 Mar 2008 19:43:56 IST
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yeah 25927 is right
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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11 Mar 2008 19:52:22 IST
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plz post ur solution netkid07
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
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11 Mar 2008 19:54:39 IST
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There are  such numbers.
Let us take an example of a four-digit number having digits a, b, c, d. Then the condition is  where only 'a' can't take 0 as its value. Hence the required number of solutions will be the coefficient of  in
 .
The coefficient comes out to be  .
Similarly repeat the procedure for two-, three-, five-, six-digit numbers. The total number of numbers between 1 and 1000000 is comes out to be 
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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11 Mar 2008 19:56:23 IST
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thanks abhijith...i was able to form the multinomial but unable to find the coefficient!
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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11 Mar 2008 19:58:35 IST
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The following formula also gives the answer:
 . You can easily obtain the answer in this manner if you look at the problem as follows: given 18 identical objects, in how many ways can we pack them into six ordered boxes, with the restriction that no more than nine objects are allowed in each box. Then,  is the number of ways this can be done without the upper limit of 9, and the  is the number of combinations that fail this condition.
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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