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 November 28th, 2016, 11:09 PM #1 Senior Member   Joined: Apr 2008 Posts: 162 Thanks: 3 a simple but time-consuming problem The problem I am going to write down is simple but very tedious. (ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from the list 1, 2, 3, 4, 5, 6, 7. Forming 3-digit numbers from 1 to 3 is easy. 123 132 213 231 312 321 Adding them up gives 1332. Things can get very ugly with 7 numbers. Is there an easier way to approach the problem? Thanks.
 November 28th, 2016, 11:23 PM #2 Global Moderator   Joined: Dec 2006 Posts: 16,604 Thanks: 1201 In your example, 1332 = 222(1 + 2 + 3). A similar calculation gives the total for each of the other 34 choices of the three digits. For example, 222(1 + 2 + 4) = 1554. Thanks from davedave
November 29th, 2016, 12:32 AM   #3
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 Originally Posted by davedave The problem I am going to write down is simple but very tedious. (ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from the list 1, 2, 3, 4, 5, 6, 7. Forming 3-digit numbers from 1 to 3 is easy. 123 132 213 231 312 321 Adding them up gives 1332. Things can get very ugly with 7 numbers. Is there an easier way to approach the problem? Thanks.
there are $7^3=343$ total 3 digit numbers using digits 1-7.

Each digit appears the same number of times as the others.

Thus each digit appears $\dfrac {343}{7} = 49$ times in the 1's, 10's, and 100's column.

Summing this we get

$T = 49(1+2+3+4+5+6+7) + 490(1+2+3+4+5+6+7)+4900(1+2+3+4+5+6+7) = (49+490+4900)(28 ) = 152292$

 November 29th, 2016, 05:58 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,409 Thanks: 861 Math Focus: Elementary mathematics and beyond As the digits are required to be different there are 210 three-digit numbers that can be formed. The required total is 93240, obtained by adapting the sum romsek gave, i.e., by using coefficients of 30, 300 and 3000. Thanks from davedave
November 29th, 2016, 08:15 AM   #5
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 Originally Posted by greg1313 As the digits are required to be different there are 210 three-digit numbers that can be formed. The required total is 93240, obtained by adapting the sum romsek gave, i.e., by using coefficients of 30, 300 and 3000.
bah, missed the word different.

shouldn't be answering these post midnight.

 November 29th, 2016, 01:48 PM #6 Senior Member   Joined: Apr 2008 Posts: 162 Thanks: 3 Thanks everyone for all your help.

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