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Mathematics 8 Online
OpenStudy (mathmath333):

question

OpenStudy (mathmath333):

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

OpenStudy (freckles):

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

OpenStudy (mathmath333):

yes

OpenStudy (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

OpenStudy (freckles):

24 things is a lot to add

OpenStudy (mathmath333):

this question should have trick

OpenStudy (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

OpenStudy (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

OpenStudy (mathmath333):

6(1+4+3+2)

OpenStudy (freckles):

yeah

OpenStudy (mathmath333):

1column =60

OpenStudy (freckles):

we will have to carry the 6

OpenStudy (mathmath333):

6+60+600+6000

OpenStudy (freckles):

and we would have 60 again but 60+6

OpenStudy (mathmath333):

6660

OpenStudy (freckles):

|dw:1440013353312:dw|

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