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Find the sum of all the four digit numbers formed using the [#permalink]
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24 Oct 2010, 11:03
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Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
I will post OA and OE tomorrow.



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Re: Nice question and a good way to solve.... [#permalink]
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24 Oct 2010, 11:22
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We can form a total of 4! or 24 numbers.When we add all these numbers ,let us look at the contribution of of the digit 2 to the sum.
When 2 occurs in the thousand place in a particular number,its contribution in the total will be 2000.the number of numbers that can be formed with 2 in the thousand place is 3! i.e 6 numbers.Hence when 2 is in the thousands place its contribution to the sum is 3! * 2000
Similarly when 2 occurs in the hundreds place, its contribution to the sum is 3! * 200
Similarly when 2 occurs in the tenth place, its contribution to the sum is 3! * 20
Similarly when 2 occurs in the unit place, its contribution to the sum is 3! * 10
The total contribution of 2 to the sum is 3! *(2000+200+20+1)=3!*2222 In a similar manner ,the contribution of 3,4, and 5 to sum will respectively be 3!*3333,3!*4444 and 3!*5555
Hence total sum using the aove four digits 3!*(2222+3333+4444+5555) i.e 3! *(2+3+4+5) * 1111
Now we can generalize the above
If all the possible n digit numbers using n distinct digits are formed ,the sum of all the numbers so formed is equal to (n1)! * ( sum of the n digits ) *( 1111...n times)
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Re: Nice question and a good way to solve.... [#permalink]
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24 Oct 2010, 11:24
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ankitranjan wrote: Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
I will post OA and OE tomorrow.
If u find this post useful ,consider giving me KUDOS. Along with the question, do post the options. if you keep the digit 2 at the one's digit. The total numbers that can be formed is 3! = 6 => 2 occurs at one's digit 6 times. Similarly all the 4 numbers occurs at one's digit 6 times. Similar all the 4 numbers occurs at all the 4 digits  one's,ten's, hundred's, thousand's 6 times. => total sum = ( 1000+100+10+1 ) *( 2+3+4+5)*6 = 1111*14*6 = 1111*84 = 93324
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Re: Nice question and a good way to solve.... [#permalink]
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24 Oct 2010, 11:33
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Re: Nice question and a good way to solve.... [#permalink]
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24 Oct 2010, 11:41
Yes understanding is very imp as pointed by bunuel. The question becomes interesting if total sum of all the 3 digits number is asked. Then the give formula is modified. the 111...n becomes 111....m where m = 3 for 3 digit number.
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Re: Nice question and a good way to solve.... [#permalink]
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02 Feb 2011, 03:52
gurpreetsingh wrote: Yes understanding is very imp as pointed by bunuel.
The question becomes interesting if total sum of all the 3 digits number is asked. Then the give formula is modified.
the 111...n becomes 111....m where m = 3 for 3 digit number. Do I get it? When you have to sum all the 3 digit numbers out of 2,3,4,5 than the answer is: without repetition: (31)!*14*111= 3108 with repetition: (3^2)*14*111=13986
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Re: Find the sum of all the four digit numbers formed using the [#permalink]
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20 Dec 2012, 05:29
How many options do I have? 4*3*2*1 How many digits will be repeated equally in the options? 4 \(=\frac{4*3*2*1}{4}*14*1111=84*1111=64*1111=93,324\)
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Re: Find the sum of all the four digit numbers formed using the [#permalink]
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27 Apr 2013, 08:21
Hi Bunuel, So in the below question if we apply the formula shouldn't we consider 4 as n? Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition. (41)*(2+3+4+5)*1111 Many thanks for your help. Aybige



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Re: Find the sum of all the four digit numbers formed using the [#permalink]
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28 Apr 2013, 03:28




Re: Find the sum of all the four digit numbers formed using the
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