## mathmath333 one year ago question

1. mathmath333

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

2. freckles

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

3. mathmath333

yes

4. 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

5. freckles

24 things is a lot to add

6. mathmath333

this question should have trick

7. 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

8. 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

9. mathmath333

6(1+4+3+2)

10. freckles

yeah

11. mathmath333

1column =60

12. freckles

we will have to carry the 6

13. mathmath333

6+60+600+6000

14. freckles

and we would have 60 again but 60+6

15. mathmath333

6660

16. freckles

|dw:1440013353312:dw|

17. freckles

hmm I got a extra 6 than you

18. mathmath333

19. mathmath333

is it 66660

20. freckles

|dw:1440013568285:dw|

21. freckles

yes it is 66660

22. freckles

|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

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

ok thns

25. 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 $12 \cdot 100+12 \cdot 10+12$