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What is the sum of all 4-digit numbers that can be formed

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What is the sum of all 4-digit numbers that can be formed [#permalink] New post 25 May 2010, 23:38
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What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

I am sorry I don't have the OA. But I think it is solvable without the OA
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Re: Can someone help? [#permalink] New post 26 May 2010, 02:32
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dimitri92 wrote:
I am sorry I don't have the OA. But I think it is solvable without the OA

What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?


As each digit of four digit number (abcd) can take 4 options (1, 2, 3, 4) total there can be 4^4 numbers.

Obviously each digit (a, b, c, and d) will take the value of 1, 2, 3, 4 equal number of times, so each digit will take the value of 1, 2, 3, 4 - \frac{4^4}{4}=4^3=64 times: units digit will take the values of 1, 2, 3, 4 - 64 times and the same with tens, hundreds, thousands digits.

So the sum would be 64*(1+2+3+4)+64*10*(1+2+3+4)+64*100*(1+2+3+4)+64*1000*(1+2+3+4)=64*10*(1+10+100+1000)=711040.

Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

1. Sum of all the numbers which can be formed by using the n digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).

2. Sum of all the numbers which can be formed by using the n digits (repetition being allowed) is: n^{n-1}*(sum of the digits)*(111…..n times).

Hope it helps.
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Re: Can someone help? [#permalink] New post 26 May 2010, 22:47
merci !! very well done
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Re: Can someone help? [#permalink] New post 26 May 2010, 22:57
Could you plz explain this :

so each digit will take the value of 1, 2, 3, 4 - \frac{4^4}{4}=4^3=64 times

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Re: Can someone help? [#permalink] New post 27 May 2010, 07:13
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sag wrote:
Could you plz explain this :

so each digit will take the value of 1, 2, 3, 4 - \frac{4^4}{4}=4^3=64 times

Thanks & Regards


Total such numbers = 4^4 = 256.

1/4 of these numbers, or 64 numbers, will have units digit of 1; another 1/4 will have the units digit of 2; another 1/4 will have the units digit of 3; and the last 1/4 will have the units digit of 4.

The same with tens, hundreds, thousands digits.

Hope it's clear.
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Re: Can someone help? [#permalink] New post 27 May 2010, 21:41
Thanks Bunuel.. +1..
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Re: Can someone help? [#permalink] New post 23 Aug 2010, 06:43
Agreed. These are powerful formulas but the concept behind it should be understood before roting them.
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Re: What is the sum of all 4-digit numbers that can be formed [#permalink] New post 14 Oct 2013, 12:41
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Re: What is the sum of all 4-digit numbers that can be formed   [#permalink] 14 Oct 2013, 12:41
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