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OpenStudy (mathmath333):
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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
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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
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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
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OpenStudy (freckles):
|dw:1440013353312:dw|
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