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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Nov 2006 04:29:21 IST
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hi
plz solve this question
" find the sum of all 5-digit numbers that can be formed using the digits 1,2,3,4,5 if repetition is not allowed."
need an urgent reply!!!!!!!!
ruhi
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16 Nov 2006 04:44:35 IST
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studying so early in the morn!!!  !!!!
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16 Nov 2006 04:45:32 IST
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yaa
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16 Nov 2006 09:36:35 IST
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As choosing a set of 5 out of {1,2,3,4,5} is possible in only one way
12345
So number of all different numbers can be formed by this set is 5! = 120
Now imagine placing each of these 5-digit numbers on top of each other in a long list to be added manually. Each of the digits 1, 2, 3, 4, and 5 will appear equally often in each of the units, tens, hundred, thousand and ten thousand columns. There are two ways to proceed...
1.
As 120/5 = 24, each digit will contain twenty-four occurrences of each digit and so each column would add to 24(1+2+3+4+5) = 360.
In adding the units column we write 0 and carry 36.
In the ten column we get 360+36 = 396: write 6 and carry 39.
In the hundred column we get 360+39 = 399: write 9 and carry 39.
In the thousand column we get 360+39 = 399: write 9 and carry 39.
In the ten thousand column we get 360+39 = 399: write 9 and carry 39.
Hence the sum is 39 9 9 9 6 0 = 3,999,960.
2.
Its a tricky one
As the mean digit in each column is 3, each number is 33333, on average. Hence the sum is 120 33333= 3,999,960.
Now you can try to solve this one
Find the sum of all possible permutations of k digits taken from {1,2,3,...,n}, if the repetition is not allowed.
Cheers
Rahul
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Rahul A
-- If you thing you can, you can. If you think you can't, you're right !! -- |
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16 Nov 2006 10:17:43 IST
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hey rahul
i like the 2nd method yaar. thats a cool method. thanx 4 sharing it with me.
ruhi
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16 Nov 2006 10:30:23 IST
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Rahul A
-- If you thing you can, you can. If you think you can't, you're right !! -- |
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16 Nov 2006 13:20:25 IST
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Gr8 answer ... rahul gr8 work done .. cheers
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Puneet Agrawal
IIT Delhi
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16 Nov 2006 18:16:15 IST
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hey rahul , i liked the second method. But if there are 6 digits ( as in 1,2,3,4,5,6) , then what? is this method still applicable? Sharry
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16 Nov 2006 18:57:14 IST
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i think the 2nd method is applicable only 4 odd no. of digits nd 4 even no. of digits 1st method will work which is also a good one
ruhi
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17 Nov 2006 09:26:10 IST
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Well I have not tried, but here is a technique, may be i am wrong
For 6 numbers, total possible numbers are 6! = 720
take half of the the average numbers = 333,333 and other half = 444,444
So the sum can be S = 360 x 333,333 + 360 x 444,444
Try to find out the solution with the first method and compare, may be you get both the answers matched
Cheers
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Rahul A
-- If you thing you can, you can. If you think you can't, you're right !! -- |
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22 Nov 2006 20:34:52 IST
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wow ! hey rahul ....u're a mathematician yaar. cool!! thnx.
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