Question

Find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time!!

Answer 1

@Everyone

Answer 2

Count repeats?

Answer 3

@Ashleyisakitty @april115 @e.cociuba @thomaster @thomaster @Loujoelou @countonme123 @Muddingirl @MathFan12

Answer 4

Yes

Answer 5

So, 5555 would be a number

Answer 6

@dimensionx @skullpatrol @satellite73

Answer 7

Yes

Answer 8

@KingGeorge @karatechopper

Answer 9

First though is to write it as \[\sum_{i=2}^5\sum_{j=2}^5\sum_{k=2}^5\sum_{l=2}^5 (i\cdot 10^3+j\cdot10^2+k\cdot10+l)\]

Answer 10

kay

Answer 11

@KingGeorge Can u refer any good websites for permutations and combinations...........M kinda new to the topic!!

Answer 12

I don't know of any good websites off the top of my head. But try searching for stuff about "combinatorics." This might get you some stuff you're interested in as long as you filter out some of the more advanced material.

Answer 13

I don't know of any good ways to simplify the sum I wrote, but any decent graphing calculator or wolfram alpha should be able to sum that.

I'll keep thinking to see if it can be simplified at all though.

Answer 14

ta

Answer 15

Are you sure that you can repeat digits?

Answer 16

In any case, I did find a better solution. One minute while I type up a clear explanation.

Answer 17

yes.sure........think I got it............The ans is 93324??

Answer 18

Out of all of the 4-digit numbers you make, each one will have an equal probability of have a 2 in the first place, as it will a 5 or a 3 or a 4. And this is the same for the second/third/fourth places as well. So we can look at the average number that will be in these places. As it turns out, this will be \((2+3+4+5)/4=14/4=7/2\).

Now, exactly how many numbers will there be total? We have 4 choices for each place, and 4 places. Thus, there are \(4^4=256\) different possibilities. So if we look at only a specific digit, the sum will be \(256*(7/2)=896\).

Finally, we look at the different places. This results in the formula\[896*1000+896*100+896*10+896\]Run that through a calculator, and you should have an answer.

And no, it is not 93324.

Answer 19

I think Its 4! ways * 7/2 ..not 4^4

Answer 20

We're allowing repetition, so it's 4^4. If repetition is not allowed, then yes, it would be 4!

Answer 21

permutation so \[4P _{4}=4!\]

Answer 22

Check to make absolute sure that you're including repetitions. Can you type out the exact wording here? Or did you already do that at the top?

Answer 23

http://answers.yahoo.com/question/index?qid=20091115020541AA4ww3V
I alredy did dat.......I found this and this matches with 93324

Answer 24

That problem specifies that there are no repeats (that's what the "by permutation" means).

Answer 25

I mentioned 4 comments ago,its permutation so 4!

Answer 26

If you are indeed instructed to use permutations, then yes, the solution in that link is correct.

Answer 27

so help me with this if it was not mentioned as permutations!!,we always shud take repetition into account??

Answer 28

Correct. If the problem is as it was stated at the very top, then you should consider repetitions.

Answer 29

But it is the same I typed!!!,given under permutation section!!

Answer 30

Can you do me a favor and please type the problem EXACTLY as it was written?

Answer 31

That is the xact question!!!

Answer 32

"Find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time?"

Answer 33

yes..........King!!

Answer 34

If that's the exact phrasing, then I think you need to consider repetitions.

Answer 35

O'cmon..........Even if its given under permutation section,need I consider repetitions??

Answer 36

dats a gen question,btw!!

Answer 37

If it's in the permutation section, then maybe you should use permutations, in which case repetitions wouldn't be considered.

Go ahead and use the permutations then. At worst, you should still get most of the credit for having the right method.

Answer 38

guess wat.......Its 1 point question!!

Answer 39

Go with the permutations then.

Answer 40

And to think I just hav to waste dat long time for a 1pointer!!!O_O

Answer 41

Anyways..........tnx for the help O'King!!Appreciate it!!

Answer 42

No problem. Sorry about all that confusion :P

Answer 43

It still shows I can bump it after 10010 mins........o_O.......
NP bout dat!!

Answer 44

Np on the confusion!!Happens!!Like it is now with me and the website timer!!