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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    Using the digits \(1,2,3,4\) find sum of all possible 4 digit number without repetition.

  2. freckles
    • one year ago
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    example: pretend we just had 1,2,3 instead... are they saying they want: 123+132+213+231+312+321

  3. mathmath333
    • one year ago
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    yes

  4. freckles
    • one year ago
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    so I would do the same thing there like start off with the 1 and do rearrangements of the others we should wind up with adding 4(3)(2)(1) arrangements

  5. freckles
    • one year ago
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    24 things is a lot to add

  6. mathmath333
    • one year ago
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    this question should have trick

  7. freckles
    • one year ago
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    1234 1243 1324 1342 1432 1423 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4312 4321 4213 4231 4123 4132

  8. freckles
    • one year ago
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    so for the one's column we would be adding 6 amount of 4's and 6 amount of 3's and 6 amount of 2's and 6 amount of 1's and so on for the other columns

  9. mathmath333
    • one year ago
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    6(1+4+3+2)

  10. freckles
    • one year ago
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    yeah

  11. mathmath333
    • one year ago
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    1column =60

  12. freckles
    • one year ago
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    we will have to carry the 6

  13. mathmath333
    • one year ago
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    6+60+600+6000

  14. freckles
    • one year ago
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    and we would have 60 again but 60+6

  15. mathmath333
    • one year ago
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    6660

  16. freckles
    • one year ago
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    |dw:1440013353312:dw|

  17. freckles
    • one year ago
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    hmm I got a extra 6 than you

  18. mathmath333
    • one year ago
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    ur answer is correct ?

  19. mathmath333
    • one year ago
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    is it 66660

  20. freckles
    • one year ago
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    |dw:1440013568285:dw|

  21. freckles
    • one year ago
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    yes it is 66660

  22. freckles
    • one year ago
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    |dw:1440013677810:dw| also when I wrote this I hope you know that (60)(60)(60)(60) wasn't meant to be multiplication

  23. freckles
    • one year ago
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    60 was in the ones digit 60 was in the tens digit 60 was in the hundreds digit 60 was in the thousands digit

  24. mathmath333
    • one year ago
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    ok thns

  25. freckles
    • one year ago
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    guess we could have done the same thing for 123+132+213+231+312+321 ...we had 1,2,3 there were going to be 6 arrangements there are 3 numbers so 6/3=2 so there was going to be 2(3+2+1) for each column which is 12 so we would have the number (12)(12)(12) <---not multiplication; there are digits \[12 \cdot 100+12 \cdot 10+12\]

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