Let's do some case work. One-digit numbers: There are four such numbers: $1, 2, 3, 4$. They add to $10$. Two-digit numbers: Since repetition is not permitted, there are four ways to choose the tens digit and three ways to choose the units digit. Hence, there are $4 \cdot 3 = 12$ such numbers. By symmetry, each of the digits $1, 2, 3, 4$ must appear in each position $12/4 = 3$ times. Hence, the sum of the two-digit numbers which can be formed from the digits $1, 2, 3, 4$ without repetition is $$\frac{1}{4} \cdot 4 \cdot 3 \cdot (1 + 2 + 3 + 4)(10 + 1) = 330$$ Three-digit numbers: Since repetition is not permitted, for each of the four choices for the hundreds digit, there are three choices for the tens digit, and two choices for the units digit. Hence, there are $4 \cdot 3 \cdot 2 = 24$ such numbers. By symmetry, each of the digits $1, 2, 3, 4$ must appear in each position $24/4 = 6$ times. Hence, the sum of the three-digit numbers which can be formed from the digits $1, 2, 3, 4$ without repetition is Four-digit numbers: Since repetition is not permitted, for each of the four choices for the thousands digit, there are three choices for the hundreds digit, two choices for the tens digit, and one choice for the units digit. Hence, there are $4 \cdot 3 \cdot 2 \cdot 1 = 24$ such numbers. By symmetry, each of the digits $1, 2, 3, 4$ must appear in each position $24/4 = 6$ times. Hence, the sum of the four-digit numbers which can be formed from the digits $1, 2, 3, 4$ without repetition is Total: Since the four cases are mutually exclusive and exhaustive, the sum of all the numbers that can be formed using the digits $1, 2, 3,$ and $4$ without repetition is $$10 + 330 + 6660 + 66660 = 73660$$ Answer  Hint: The first thing we must find out is the total number of possible arrangements of the 4 digits of the given 5, i.e., $ ^{5}{{P}_{4}} $ . Out of these arrangements, in one fifth of the cases 5 will be present in unit place, in other one fifth cases in the tens place, in another one fifth cases in the hundreds place, one fifth in the thousands place and not present in left one fifth cases. So, we can say that each term will appear at a certain place for one fifth of the total case. So, the sum of the unit digit = $ ^{5}{{P}_{4}}\left( 1+2+3+4+5 \right) $ , Similarly, the result of tens place will also be same just ten times of the ones place. So, add all the places of the digits to get the answer. Complete step-by-step answer: The first thing we must find out is the total number of possible arrangements of the 4 digits of the given 5, as the number of ways in which 4 out of the given 5 digits can be arranged is the number of 4 digit numbers formed. Therefore, the number of 4 digit numbers formed from the given 5 digits is $ ^{5}{{P}_{4}} $.So, we are sure that each of the places of the numbers, i.e., ones place, tens place, hundreds place and thousands place have \[\dfrac{^{5}{{P}_{4}}}{5}\] number of times a digit appearing.So, the sum of ones digit of all the numbers is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right) $ Similarly, tens digit will have the same sum, but we need to multiply the result by 10, as the place value of tens place is ten. So, the value of tens place addition is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 10 $ Similarly, the 100s place value is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 100 $ and 1000s place value is $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 1000. $ Now we will add the place values of all the places to get the answer.Therefore, the sum of all the numbers satisfying the above condition is:\[\begin{align} & \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)+\dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 10+ \\ & \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 100+\dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 1000 \\ \end{align}\] $ =\dfrac{^{5}{{P}_{4}}}{5}\times 15{{+}^{5}}{{P}_{4}}\times 15\times 10+\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 100+\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1000 $ $ =\dfrac{^{5}{{P}_{4}}}{5}\times 15\left( 1+10+100+1000 \right) $ $ =\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1111 $ Now we know that $ ^{5}{{P}_{4}}=\dfrac{5!}{1!}=120 $ .\[\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1111=\dfrac{120}{5}\times 15\times 1111=399960\]Therefore, the answer to the above question is 399960.Note: Remember if one of the digits is zero, then all the digits will not have the same cases of appearing in a given place, actually zero cannot appear at the 1000s place, as if it appears at 1000s place, the number will be a 3 digit number. So, in this case you have to fix zero at every place and get the number of times zero is appearing at each place and divide the left out cases among other digits.
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What is the sum of all 4-digit numbers that can be formed [#permalink]
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Question Stats: Hide Show timer StatisticsWhat 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
Math Expert Joined: 02 Sep 2009 Posts: 86743
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
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. _________________
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
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.[/quote]Hi Bunuel,Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong.[/quote] Yes, the numbers range from 1111 to 4444: 1111, 1112, 1113, 1114, 1121, 1121, ..., 4444. But the numbers are NOT evenly spaced so you cannot get the sum by multiplying the mean by the number of terms.Does this make sense?[/quote]I am no Math Expert. Can you please correct my below approach?We know the smallest number we can make is 1111 and the largest number we canmake is 4444.We also know that our numbers will be evenly distributed in the middle (i. e. 1112 isbalanced by 4443; 1113 is balanced by 4442). So, we can solve using the averageformula.Finally, we know that there are 4*4*4*4 = 256 numbers in our set.Average = sum of terms/# of termssum of terms = average * # of termssum of terms = (1111+4444)/2 * 256 = 5555/2 * 256 = 5555*128 = 711040
Manager Joined: 15 Nov 2006 Affiliations: SPG Posts: 242
Re: What is the sum of all 4-digit numbers that can be formed [#permalink] merci !! very well done
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Re: What is the sum of all 4-digit numbers that can be formed [#permalink] 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
Math Expert Joined: 02 Sep 2009 Posts: 86743
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
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: What is the sum of all 4-digit numbers that can be formed [#permalink] Thanks Bunuel.. +1..
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Re: What is the sum of all 4-digit numbers that can be formed [#permalink] Agreed. These are powerful formulas but the concept behind it should be understood before roting them.
Intern Joined: 05 Feb 2014 Posts: 35
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
Bunuel wrote: 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. Hi Bunuel,Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong.
Intern Joined: 13 May 2014 Posts: 28 Concentration: General Management, Strategy
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
gauravsoni wrote: Bunuel wrote: 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. Hi Bunuel,Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong. The range of the numbers do vary from 1111 to 4444 inclusive, and there are only 264 different numbers altogether formed.however, what do you mean by the sum of the numbers will be 1111 to 4444 is not clear. The formula as given when repetition is allowed is pretty simple :n^{n-1}*(sum of the digits)*(111…..n times). Here, n^{n-1} : the no of times each digit appear at each placeneeds to be multiplied by sum (of the digits) as all the digits take that placemultiplied by 111... upto n times - to finally find the value at each placeYou could see the manner in which the total sum could be arrived at with the formula.Hope it helps.Press kudos if you wish to appreciate
Intern Joined: 16 May 2014 Posts: 32
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
gauravsoni wrote: Bunuel wrote: 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. Hi Bunuel,Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong. All the numbers which are formed lies between 1111 and 4444 but it does not include all numbers from 1111 to 4444. For example, 1235 will not be formed as we have only 1,2,3,4 to choose from. Thus, we can't use the formula of arithmetic mean. Hope it clears your doubt!!Kudos if it does!!!
Math Expert Joined: 02 Sep 2009 Posts: 86743
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
gauravsoni wrote: Bunuel wrote: 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. Hi Bunuel,Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong. Yes, the numbers range from 1111 to 4444: 1111, 1112, 1113, 1114, 1121, 1121, ..., 4444. But the numbers are NOT evenly spaced so you cannot get the sum by multiplying the mean by the number of terms.Does this make sense? _________________
Intern Joined: 05 Feb 2014 Posts: 35
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
Bunuel wrote: Bunuel wrote: 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. Hi Bunuel,Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong. Yes, the numbers range from 1111 to 4444: 1111, 1112, 1113, 1114, 1121, 1121, ..., 4444. But the numbers are NOT evenly spaced so you cannot get the sum by multiplying the mean by the number of terms.Does this make sense?[/quote]Ah yes , got it thanks.
Intern Joined: 28 Apr 2015 Posts: 1 GMAT 1: 570 Q43 V25 GMAT 2: 700 Q46 V40
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
Bunuel wrote: 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. Did you mean "1111"? (highlighted in red)
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Re: What is the sum of all 4-digit numbers that can be formed [#permalink] Hi All,While much of this discussion is over a year old, it's important to note how important it is to include the 5 answer choices to any PS question. The GMAT only rarely offers questions that can only be solved by 'doing math in one specific way', which means that there are normally several different ways to approach each question. By having the answer choices to work with, we can sometimes avoid doing math altogether (since can use estimation or logic to determine that certain answers are 'too small' or 'too big' to be correct).Here, by not including the 5 answer choices, the original poster forces us to do math, when a more elegant, simpler or faster approach might have been possible.GMAT assassins aren't born, they're made,Rich _________________
Re: What is the sum of all 4-digit numbers that can be formed [#permalink] Here is an alternative solution that you can compute easily on the GMAT if given this question:First, note that the smallest number you can make is 1111, the largest is 4444. It is reasonable to conclude that the average of all the combinations of digits (1,2,3,4) very close to the average of 1111+4444, which is 5555/2.Next, compute the possibilities: 4*4*4*4 = 16*16 = 256 (memorize this). 4*4*4*4. Now all you have to do is compute 256 * 5555 / 2.
Intern Joined: 13 Jan 2017 Posts: 2
Re: What is the sum of all 4-digit numbers that can be formed [#permalink] At 1000nds place 4,3 2,1 can all come likewise at 100th ,10th and unit place also all of it can come
Current Student Joined: 05 Jul 2018 Status:Current student at IIMB Affiliations: IIM Bangalore Posts: 400 Location: India Concentration: General Management, Technology Schools: IIMB EPGP'22 (A) GMAT 1: 600 Q47 V26 GRE 1: Q162 V149
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Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
Bunuel wrote: 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? 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. Bunuel: Has there ever been such a question on official GMAT...If so please share some reference link.
Math Expert Joined: 02 Sep 2009 Posts: 86743
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
saukrit wrote: Bunuel wrote: 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? 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. Bunuel: Has there ever been such a question on official GMAT...If so please share some reference link. Check here: Constructing Numbers, Codes and Passwords. _________________
VP Joined: 07 Dec 2014 Posts: 1134
What is the sum of all 4-digit numbers that can be formed [#permalink]
dimitri92 wrote: 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? least possible value=1111greatest possible value=44441111+4444=55555555/2=2777.5=meanbecause each of 4 digits can take any of 4 options,sum=4^4*2777.5=711040
What is the sum of all 4-digit numbers that can be formed [#permalink] 12 Jun 2019, 19:49 |
What is the sum of all 4 digit numbers that can be formed using all of the digits 2 3 4 6 without repetition?
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