How many 3 digit numbers can be formed from the numbers 1,2 and 3 if all the three digits must be different must be unique?
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This section covers permutations and combinations. Show Arranging Objects The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1 Example How many different ways can the letters P, Q, R, S be arranged? The answer is 4! = 24. This is because there are four spaces to be filled: _, _, _, _ The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!
n!
. Example In how many ways can the letters in the word: STATISTICS be arranged? There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are: 10!=50 400 Rings and Roundabouts
When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)! Example Ten people go to a party. How many different ways can they be seated? Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440 Combinations The number of ways of selecting r objects from n unlike objects is:  Example There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls? 10C3 =10!=10 × 9 × 8= 120 Permutations A permutation is an ordered arrangement.
nPr = n! . Example In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use. 10P3 =10! = 720 There are therefore 720 different ways of picking the top three goals. Probability The above facts can be used to help solve problems in probability. Example In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery? The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 . Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance. GMAT Club Daily PrepThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.Customized we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice we will pick new questions that match your level based on your Timer History Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.Hello Guest!It appears that you are browsing the GMAT Club forum unregistered! Signing up is free, quick, and confidential. Join 700,000+ members and get the full benefits of GMAT ClubRegistration gives you:
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Intern Joined: 02 Jan 2011 Posts: 2 There are 27 different three-digit integers that can be formed using [#permalink]
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Hide Show timer StatisticsThere are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be? A. 2,704 Originally posted by rdhookar on 13 Jan 2011, 18:58. Renamed the topic, edited the question and added the OA. GMAT Club Legend Joined: 08 Jul 2010 Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator Posts: 5905 Location: India GMAT: QUANT EXPERT WE:Education (Education)
Re: There are 27 different three-digit integers that can be formed using [#permalink]
rdhookar wrote: There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be? A. 2,704 Since every digit has equal likely chances of appearing at any of the three places in a 3 digit number so out of 27 number 9 will have Unit digit 1, 9 will have unit digit 2 and 9 will have unit digit 3 So sum of the unit digits of all 27 numbers = 9(1+2+3) = 54 We write 4 at the unit digit place and 5 goes carry forward so out of 27 number 9 will have Tens digit 1, 9 will have Tens digit 2 and 9 will have Tens digit 3 So sum of the Tens digits of all 27 numbers = 9(1+2+3) = 54 but 5 is carry forward so the sum becomes 54+5 = 59, We write 9 at the tens digit place and 5 goes carry forward This much calculation gives us the answer out of given options however we can do the same thing two more times to find exact sum which will be 5,994 Answer: option E GMATinsight Manager Joined: 02 Apr 2010 Posts: 81 Re: There are 27 different three-digit integers that can be formed using [#permalink]
Probably the easiest and fastest way to answer this problem is to think about the average three digit number you can create using just the digits 1,2,3, and to multiply this average number with 27. The average three digit number is 222 (choosing an average of 1,2,3 for every single digit of the number). So 222*27 = 5994. Hence, solution A is correct. Intern Joined: 24 Jun 2010 Posts: 10 Re: There are 27 different three-digit integers that can be formed using [#permalink]
here you go: 27 digits are formed by 3 * 3 *3 that is..1st place can take 1,2,3...again 2nd place can take all three digits...3rd place can also take all three digits. now lets consider the last digit as 1 out of the three that we can take: 3*3*(1) so with 1 as the last digit we can have 9 combinations.so we have 9 1's in the last place similary,lets consider the last digit as 2 out of the three that we can take: 3*3*(2) so with 2 as the last digit we can have 9 combinations.so we have 9 2's in the last place similary,lets consider the last digit as 3 out of the three that we can take: 3*3*(3) so with 3 as the last digit we can have 9 combinations.so we have 9 3's in the last place therefore sum of 9 1's,9 2's,9 3's will be 9+18+27=54...similarly 2nd and 1st place will give u 54...therefore 5994..hope u got it!! Manager Joined: 13 Jun 2012 Posts: 177 Location: United States WE:Supply Chain Management (Computer Hardware)
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The simplest way is this formula n^n-1 *sum of the numbers (111)= 5994. where n= total number, which is 3 here. so 3^3-1=9. 9+6(111)=5994 Manager Joined: 10 Mar 2013 Posts: 145 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 GPA: 3
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There are 27 different three-digit integers that can be formed using [#permalink]
swaraj wrote: here you go: 27 digits are formed by 3 * 3 *3 that is..1st place can take 1,2,3...again 2nd place can take all three digits...3rd place can also take all three digits. now lets consider the last digit as 1 out of the three that we can take: 3*3*(1) so with 1 as the last digit we can have 9 combinations.so we have 9 1's in the last place similary,lets consider the last digit as 2 out of the three that we can take: 3*3*(2) so with 2 as the last digit we can have 9 combinations.so we have 9 2's in the last place similary,lets consider the last digit as 3 out of the three that we can take: 3*3*(3) so with 3 as the last digit we can have 9 combinations.so we have 9 3's in the last place therefore sum of 9 1's,9 2's,9 3's will be 9+18+27=54...similarly 2nd and 1st place will give u 54...therefore 5994..hope u got it!! This is a great explanation! Thanks! Intern Joined: 18 Dec 2015 Posts: 1 GPA: 3.15 WE:Analyst (Non-Profit and Government) Re: There are 27 different three-digit integers that can be formed using [#permalink]
How I did it: 123 is a possible solution, which is divisible by 3. E is the only option that adds up to a number divisible by 3. EMPOWERgmat Instructor Joined: 19 Dec 2014 Status:GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Posts: 20932 Location: United States (CA) Re: There are 27 different three-digit integers that can be formed using
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Hi All, This question has a great series of built-in 'logic shortcuts' that don't require a lot of advanced math (although a bit of arithmetic will still be required to get to the correct answer). We're told that there are 27 3-digit numbers that can be formed with the digits 1, 2 and 3 (the smallest being 111 and the largest being 333). Without too much effort, you should be able to figure out that... 9 of the numbers begin with a '1' So when we add up JUST the 100s, we have.... 9(100) + 9(200) + 9(300) = 5400 Next, consider ALL of the 10s that we have... 27 of them (and some are 20s and some are 30s). That would add HUNDREDS (well over 270+) to the total sum. Based on how the answer choices are written, there's only one answer that makes sense... Final Answer: GMAT assassins aren't born, they're made, Target Test Prep Representative Joined: 14 Oct 2015 Status:Founder & CEO Affiliations: Target Test Prep Posts: 16224 Location: United States (CA) Re: There are 27 different three-digit integers that can be formed using [#permalink]
rdhookar wrote: There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be? A. 2,704 Some of the integers that have this property are 123, 111, 213, and 322. Of course, it’s possible to list all 27 such integers and then add them up. However, it will be too time-consuming. Therefore, we will use a shortcut by arguing that, of these 27 integers, the digits 1, 2, and 3 each must appear in the hundreds position 9 times. Using the same logic, they each will also appear in the tens position 9 times and in the units position 9 times. Thus, the sum of these integers is: (100 + 200 + 300) x 9 + (10 + 20 + 30) x 9 + (1 + 2 + 3) x 9 (600) x 9 + (60) x 9 + (6) x 9 (666) x 9 5,994 Answer: E See why Target Test Prep is the top rated GMAT course on GMAT Club. Read Our Reviews Manager Joined: 30 Jul 2014 Status:MBA Completed Affiliations: IIM C, IIM A Posts: 92 Location: India
Re: There are 27 different three-digit integers that can be formed using [#permalink]
Sum of the numbers should be divisible by 9 as each number is appearing 9 times. Divisibility rule by 9 conforms(Sum of all
the digits should be divisible by 9) only to the answer option E - hence answer should be E. For application and interview help, feel free to email me - Manager Joined: 01 Dec 2018 Posts: 176 Concentration: Entrepreneurship, Finance GPA: 4 WE:Other (Retail Banking) Re: There are 27 different three-digit integers that can be formed using [#permalink]
rdhookar wrote: There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be? A. 2,704 SUPER EASY METHOD that always works STEP 1 always find the total no. of combinations possible = here we have been given 27 (use fundamental method to check/find 3x3x3 =27) Senior Manager Joined: 26 Apr 2019 Posts: 251 Location: India Concentration: Finance, Leadership GMAT 1: 690 Q49 V34 GMAT 2: 700 Q49 V36 GMAT 3: 720 Q50 V37 GMAT 4: 740 Q50 V40 GPA: 3.99
Re: There are 27 different three-digit integers that can be formed using
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at ones place there will be nine 1 , nine 2 and nine 3 , adding all 27 will give you 54 . silmilarly you will get 54 at tens place and 54 at hundred place . all all we get e as answer VP Joined: 07 Dec 2014 Posts: 1133 Re: There are 27 different three-digit integers that can be formed using
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rdhookar wrote: There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be? A. 2,704 111+333=444 Intern Joined: 20 Jul 2020 Posts: 6 Re: There are 27 different three-digit integers that can be
formed using [#permalink]
I think there is a simple formula that can be applied here. My only confusion is regarding how we can figure out whether or not the question implies repetition of numbers. The formula without the repetition gives the answer of course, but can we determine that for sure from the question? EMPOWERgmat Instructor Joined: 19 Dec 2014 Status:GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Posts: 20932 Location: United States (CA) Re: There are 27 different three-digit
integers that can be formed using [#permalink]
salphonso wrote: I think there is a simple formula that can be applied here. My only confusion is regarding how we can figure out whether or not the question implies repetition of numbers. The formula without the repetition gives the answer of course, but can we determine that for sure from the question? Hi salphonso, The prompt specifically tells us that we're dealing with "27 DIFFERENT three-digit integers...", so we know that the numbers do NOT repeat. GMAT questions are always carefully worded, so you should pay attention to the specific 'math vocabulary' that is used in each prompt (as that can help you to properly 'restrict' your thinking or get you thinking about possibilities that are not immediately obvious to you). GMAT assassins aren't born, they're made, Intern Joined: 20 Jul 2020 Posts: 6 Re: There are
27 different three-digit integers that can be formed using [#permalink]
Hi Rich, Can't believe I missed that! Thank you! Intern Joined: 18 Jul 2017 Posts: 12 There are 27 different three-digit integers that can be formed using [#permalink]
Can we use this method? 1 number would be 111 and last would be 333. Bunuel Posted from my mobile device GMAT Club Legend Joined: 11 Sep 2015 Posts: 6812 Location: Canada
Re: There are 27 different three-digit integers that can be formed using
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rdhookar wrote: There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be? A. 2,704 This is a great example of how the GMAT often rewards students for thinking outside the box. Here, we can apply a divisibility rule that says integer N is divisible by 3 if and only if the sum of the digits of N is divisible by 3. Notice that
1+2+3 = 6, and 6 is divisible by 3 Let's check each answer choice... By the process of elimination, the correct answer is E. Brent Hanneson – Creator of gmatprepnow.com Re: There are 27 different three-digit integers that can be formed using [#permalink] 28 Nov 2021, 07:55 Moderators: Senior Moderator - Masters Forum 3086 posts How many 3So I take some particular numbers, like 1,2,3 and say that, well, 1 can go in 3 places, 2 in 2 places and 3 in 1 place, so by multiplication principle, there are 6 ways of forming a 3-digit number with 1,2,3. But there are 4 different numbers. So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4).
How many 3Hence, the number of 3-digit odd numbers that can be formed =3×6×6=108.
How many numbers can you make with the numbers 1 2 and 3?That is a total of 7 combinations.
How many groups of 3Interpreting the result
There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed. Note: We can also use the multiplication principle to answer this question.
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