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© 2022 Tardigrade®. All rights reserved 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.
In this answer I presume that all digits $1,2,5,6$ must be present in the number. There are $4!=24$ such numbers. We can index them with $i=1,\dots,24$ and write each of them as $$n_{i}=a_{i}+10b_{i}+100c_{i}+1000d_{i}$$ where $\left\{ a_{i},b_{i},c_{i},d_{i}\right\}=\{1,2,5,6\} $ Then $$\sum_{i=1}^{24}n_{i}=\sum_{i=1}^{24}a_{i}+10\sum_{i=1}^{24}b_{i}+100\sum_{i=1}^{24}c_{i}+1000\sum_{i=1}^{24}d_{i}$$ $\left\{ i\mid a_{i}=1\right\} $, $\left\{ i\mid a_{i}=2\right\} $, $\left\{ i\mid a_{i}=5\right\} $ and $\left\{ i\mid a_{i}=6\right\} $ have equal cardinality $\frac{24}4=6$ so: $$\sum_{i=1}^{24}a_{i}=6.1+6.2+6.5+6.6=6(1+2+5+6)=84$$ This can also be applied for $b,c$ and $d$ and finally we find: $$\sum_{i=1}^{24}n_{i}=84+10.84+100.84+1000.84=1111.84=93324$$ Does this really answers your question? If not, then please let me know.
What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
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Question Stats: Hide Show timer StatisticsWhat is the sum of all the 4 digit numbers formed by digits 2, 3, 4, 5 if repetition is not allowed??A. 103324B. 93324C. 87645D. 77678 E. 65432 _________________
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Originally posted by fitzpratik on 29 Sep 2018, 19:31. Renamed the topic and edited the question.
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A. 103324B. 93324C. 87645D. 77678 E. 65432 add least and greatest: 2345+5432=77777777/2=3888.5=average4! possible combinations=2424*3888.5=93324B
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A. 103324B. 93324C. 87645D. 77678 E. 65432 let us take 2 as the thousands digit the following combinations are possible 234524352354253424532543let us take sum of units digit2*5 + 2*4 + 2*3 = 10+8+6 = 24similarly when we take 3 as the thousands digit 2*5 + 2*2 + 2*4 = 10+4+8=22when we take 4 as the thousands digit 2*5 + 2*2 + 2*3 = 10+4+6=20when we take 5 as the thousands digit2*2 + 2*3 + 2*4 = 4+6+8=18sum of units digit = 84similarly sum of tens digit = 84*10sum of hundreds digit =84*100sum of thousands digit = 84*1000sum total = 84+84*10+84*100+84*1000 = 84(1111) = 93324B
Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink] The way I solved it was using the formula suggested by bunuelon the following link: https://gmatclub.com/forum/what-is-the- ... 94836.html _________________
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink] I solved this question using another approach, Posted from my mobile device
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A. 103324B. 93324C. 87645D. 77678 E. 65432 carcass gmatbusters VeritasKarishma Dear Experts,Kindly help with a solution on this one please. Thank you.
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink] Four digit numbers =4⋅3⋅2⋅1=24 ways we can form a four digit number. Since it's a 4 digit number, each digit will appear 6=24/4 times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is 6(2+3+4+5)=85Why they are adding 2+3+4+5?= Because each digit will come at those places.Similarly sum of digits at tens,hundreds,thousands place is 84Required sum of all numbers =84+84⋅10+84⋅100+84⋅1000=93324 poojasingavi wrote: fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A. 103324B. 93324C. 87645D. 77678 E. 65432 carcass gmatbusters VeritasKarishma Dear Experts,Kindly help with a solution on this one please. Thank you. _________________
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
poojasingavi wrote: fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A. 103324B. 93324C. 87645D. 77678 E. 65432 carcass gmatbusters VeritasKarishma Dear Experts,Kindly help with a solution on this one please. Thank you. Request to all forum users: I am one of the forum experts on the PS and CR forums. If you would like me to take a look at any question on these forums, please mention it on my thread (link given below):PS - https://gmatclub.com/forum/veritas-prep ... 78042.html CR - https://gmatclub.com/forum/veritas-prep ... 78026.htmlSince I receive 100s of mails everyday in my mailbox, unfortunately, it becomes very difficult to follow up on every query even if I am tagged on it. I do take a look at both my threads on a daily basis and hence will certainly get to your query if you mention on it. Thank you!Coming to the question:What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A few things to note here: The numbers will be of the form 2354, 3425 ... etcThere will be a total of 4*3*2*1 = 24 numbersThese will be made symmetrically. What this means is that in the 24 numbers, 2 will appear 6 times, 3 will appear 6 times, 4 will appear 6 times and 5 will appear 6 times (since all digits are equivalent)23453245...23543254...etcThis will be true of digits in tens place, hundreds place and thousands place too.The sum of digits in units place = 6*2 + 6*3 + 6*4 + 6*5 = 6*(2 + 3 + 4 + 5) = 84The sum of digits in tens places, hundreds places and thousands places will also be 84 each.The sum of all numbers then will be = 84 + 84*10 + 84*100 + 84*1000 = 93324Answer (B) _________________
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A. 103324B. 93324C. 87645D. 77678 E. 65432 GMATinsight : sir is there any direct formula to solve such questions?
Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink] Hi, If we are not told can we assume that the numbers can repeat?
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
bpdulog wrote: Hi, If we are not told can we assume that the numbers can repeat? YES, No mentioned on repetition of digit means that repetition of digits is allowed. and therefore all above mentioned solutions are incorrect as they assumed that repetition of digits is not allowed.Here the solution should be smalled such number = 2222Largest number = 5555Sum of the pair of largest and smallest numbers = 2222+5555 = 7777Total Such numbers = 4*4*4*4 = 256Total suc pairs = 256/2 = 128Sum of all pairs = 7777*128 = 995456 _________________
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
GMATinsight wrote: bpdulog wrote: Hi, If we are not told can we assume that the numbers can repeat? YES, No mentioned on repetition of digit means that repetition of digits is allowed. and therefore all above mentioned solutions are incorrect as they assumed that repetition of digits is not allowed.Here the solution should be smalled such number = 2222Largest number = 5555Sum of the pair of largest and smallest numbers = 2222+5555 = 7777Total Such numbers = 4*4*4*4 = 256Total suc pairs = 256/2 = 128Sum of all pairs = 7777*128 = 995456 GMATinsight : but sir your answer is not one of the given options? also what would be the solution in case repetition of digits is not allowed..
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
GMATinsight wrote: bpdulog wrote: Hi, If we are not told can we assume that the numbers can repeat? YES, No mentioned on repetition of digit means that repetition of digits is allowed. and therefore all above mentioned solutions are incorrect as they assumed that repetition of digits is not allowed.Here the solution should be smalled such number = 2222Largest number = 5555Sum of the pair of largest and smallest numbers = 2222+5555 = 7777Total Such numbers = 4*4*4*4 = 256Total suc pairs = 256/2 = 128Sum of all pairs = 7777*128 = 995456 If numbers cnt be repeated then in that case what will be the sum?Posted from my mobile device
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
Archit3110 wrote: GMATinsight wrote: bpdulog wrote: Hi, If we are not told can we assume that the numbers can repeat? YES, No mentioned on repetition of digit means that repetition of digits is allowed. and therefore all above mentioned solutions are incorrect as they assumed that repetition of digits is not allowed.Here the solution should be smalled such number = 2222Largest number = 5555Sum of the pair of largest and smallest numbers = 2222+5555 = 7777Total Such numbers = 4*4*4*4 = 256Total suc pairs = 256/2 = 128Sum of all pairs = 7777*128 = 995456 GMATinsight : but sir your answer is not one of the given options? also what would be the solution in case repetition of digits is not allowed.. That means that either question couldn't explain properly that repetition of digits is not allowed or Options are incoorect.Probably the prior assumption is correct. _________________
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink] No of ways of arranging 4- digits = No of possible 4- digit numbers = 4! = 24.Each digit will appear for 6 times at each of thousandth, hundredth, tens and ones position.The sum of the digits at ones place = 6(2 + 3 + 4 + 5) = 84.Similarly, sum of digits at tens , hundredth, thousand place = 6(2 + 3 + 4 + 5) = 84. Sum of all possible numbers = 1000 x 84 + 100 x 84 + 10 x 84 +84 = 93324.
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5?A. 103324B. 93324C. 87645D. 77678 E. 65432 Bunuel: How do we know that in this question digits can be repeated or not
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink] Given the answer choices, the question stem should include the red portion below: fitzpratik wrote: What is the sum of all the 4-digit numbers that can be formed from the digits 2, 3, 4, and 5, if repetition is not allowed?A. 103324B. 93324C. 87645D. 77678 E. 65432 Total number of integers = number of ways to arrange the four digits = 4! = 24The middle values in the set are as follows:...3425, 3452, 3524, 3542, 4235, 4253, 4325, 4352... Notice that every colored pair sums to 7777:3542+4235 = 7777 3524+4253 = 7777 3452+4325 = 7777 3425+4352 = 7777Since the 24 integers are composed of 12 such pairs, we get:Sum of the set = 12*7777 = 93324 _________________
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink]
fitzpratik wrote: What is the sum of all the 4 digit numbers formed by digits 2, 3, 4, 5 if repetition is not allowed??A. 103324B. 93324C. 87645D. 77678 E. 65432 IMO B There's a formula for such questions n= total number of digits given =4H= highest number formed by the digitsL=lowest number formed by the digitsSUM =(H+L)*n! /2put those values Sum=24/2*(2345+5432) =93324
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Re: What is the sum of all the 4 digit numbers formed by digits 2,3,4,5? [#permalink] 20 Jul 2022, 21:43 |
What is the sum of all 4 digit numbers which can be formed with digits 2 3 4 6 without repetition?
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