mathmath333
  • mathmath333
question
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmath333
  • mathmath333
Using the digits \(1,2,3,4\) find sum of all possible 4 digit number without repetition.
freckles
  • freckles
example: pretend we just had 1,2,3 instead... are they saying they want: 123+132+213+231+312+321
mathmath333
  • mathmath333
yes

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freckles
  • freckles
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
freckles
  • freckles
24 things is a lot to add
mathmath333
  • mathmath333
this question should have trick
freckles
  • freckles
1234 1243 1324 1342 1432 1423 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4312 4321 4213 4231 4123 4132
freckles
  • freckles
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
mathmath333
  • mathmath333
6(1+4+3+2)
freckles
  • freckles
yeah
mathmath333
  • mathmath333
1column =60
freckles
  • freckles
we will have to carry the 6
mathmath333
  • mathmath333
6+60+600+6000
freckles
  • freckles
and we would have 60 again but 60+6
mathmath333
  • mathmath333
6660
freckles
  • freckles
|dw:1440013353312:dw|
freckles
  • freckles
hmm I got a extra 6 than you
mathmath333
  • mathmath333
ur answer is correct ?
mathmath333
  • mathmath333
is it 66660
freckles
  • freckles
|dw:1440013568285:dw|
freckles
  • freckles
yes it is 66660
freckles
  • freckles
|dw:1440013677810:dw| also when I wrote this I hope you know that (60)(60)(60)(60) wasn't meant to be multiplication
freckles
  • freckles
60 was in the ones digit 60 was in the tens digit 60 was in the hundreds digit 60 was in the thousands digit
mathmath333
  • mathmath333
ok thns
freckles
  • freckles
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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