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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Permutation and combination tagged by: Brent@GMATPrepNow ##### This topic has 5 expert replies and 3 member replies ## Permutation and combination The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is a.127 b.128 c.502 d.501 e.512 OA: C ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15117 messages Followed by: 1859 members Upvotes: 13060 GMAT Score: 790 parveen110 wrote: The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is a.127 b.128 c.502 d.501 e.512 OA: C This problem seems a bit too convoluted for the GMAT. Also, the GMAT would say positive integers rather than natural numbers. That said, the problem does provide some worthwhile take-aways, so here's one approach: For the digits to increase from left to right, they must be DISTINCT. For each combination of distinct digits, there is only ONE acceptable ordering: from least to greatest. For example, only one viable integer can be formed if the selected digits are 7, 1 and 3: 137. Thus, for the each of the following cases, we need to count how many COMBINATIONS of distinct digits can be formed, since each combination of distinct digits will yield exactly one viable integer. Note the following constraint: None of the digits in the following cases can be 0. The reason: Since the digits must increase from left to right, any combination that includes 0 must place 0 in the leftmost position -- not possible, since the leftmost digit of a positive integer must be at least 1. Thus, all of the following cases must be formed from the following pool of digits: 1, 2, 3, 4, 5, 6, 7, 8, 9. We should also note the following: nCr = nC(n-r). To illustrate: 6C5 = 6C(6-5) = 6C1. From 6 options, the number of ways to choose 5 = 6C5 = (6*5*4*3*2)/(5*4*3*2*1) = 6. From 6 options, the number of ways to choose 1 = 6C1 = 6. Thus: 6C5 = 6C1. 2-digit integers: Number of ways to choose 2 distinct integers from 9 options = 9C2 = (9*8)/(2*1) = 36. 3-digit integers: Number of ways to choose 3 distinct integers from 9 options = 9C3 = (9*8*7)/(3*2*1) = 84. 4-digit integers: Number of ways to choose 4 distinct integers from 9 options = 9C4 = (9*8*7*6)/(4*3*2*1) = 126. 5-digit integers: Number of ways to choose 5 distinct integers from 9 options = 9C5 = 9C4 = 126. 6-digit integers: Number of ways to choose 6 distinct integers from 9 options = 9C6 = 9C3 = 84. 7-digit integers: Number of ways to choose 7 distinct integers from 9 options = 9C7 = 9C2 = 36. 8-digit integers: Number of ways to choose 8 distinct integers from 9 options = 9C8 = 9C1 = 9. 9-digit integers: Here, there is only 1 option: 123456789. Total options = 36+84+126+126+84+36+9+1 = 502. The correct answer is C. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12677 messages Followed by: 1246 members Upvotes: 5254 GMAT Score: 770 parveen110 wrote: The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is a.127 b.128 c.502 d.501 e.512 Consider the arrangement 123456789 Now PRETEND that each digit has a lightbulb inside of it, and each lightbulb is either on or off. If the lightbulb is ON, we can see that digit, if that lightbulb is OFF, we cannot see the digit. So, for example, if we turn on the lightbulbs for digits 2, 5, 7 and 8, and turn off the remaining lightbulbs, we see the resulting number 2578 Notice that the digits are nicely arranged in ascending order. In fact, with this ON/OFF setup, every resulting number will have its digits arranged in ascending order. So, all we need to do now is count the number different ways to turn on and turn off the indivudual lightbulbs. Let's take the task of turning the lightbulbs on and off, and break it into stages. Stage 1: Determine the lightbulb situation for digit 1 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 1 in 2 ways Stage 2: Determine the lightbulb situation for digit 2 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 2 in 2 ways Stage 3: Determine the lightbulb situation for digit 3 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 3 in 2 ways . . . Stage 9: Determine the lightbulb situation for digit 9 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 9 in 2 ways By the Fundamental Counting Principle (FCP), we can complete all 9 stages (and thus create a number with its digits in ascending order) in (2)(2)(2)(2)(2)(2)(2)(2)(2) ways (= 512 ways IMPORTANT: Among the 512 possibilities, we have accidentally included some scenarios that don't meet the conditions in the question. For example, among the 512 possibilities, we have accidentally included 1 scenario where ALL of the lightbulbs are off. In this case, there is NO resulting number. Likewise, the question tells us that we must have 2 or more digits, and among the 512 possibilities, we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is on (resulting in 1, 2, 3, 4,...8, and 9). So, the number of numbers that meet the given conditions = 512 - 1 - 9 = 502 = C Cheers, Brent Aside: For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775 _________________ Brent Hanneson – Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources Last edited by Brent@GMATPrepNow on Thu Feb 13, 2014 6:30 am; edited 1 time in total GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! Senior | Next Rank: 100 Posts Joined 17 Jan 2014 Posted: 91 messages Upvotes: 7 Brent@GMATPrepNow wrote: parveen110 wrote: The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is a.127 b.128 c.502 d.501 e.512 Consider the arrangement 123456789 Now PRETEND that each digit has a lightbulb inside of it, and each lightbulb is either on or off. If the lightbulb is ON, we can see that digit, if that lightbulb is OFF, we cannot see the digit. So, for example, if we turn on the lightbulbs for digits 2, 5, 7 and 8, and turn off the remaining lightbulbs, we see the resulting number 2578 Notice that the digits are nicely arranged in ascending order. In fact, with this ON/OFF setup, every resulting number will have its digits arranged in ascending order. So, all we need to do now is count the number different ways to turn on and turn off the indivudual lightbulbs. Let's take the task of turning the lightbulbs on and off, and break it into stages. Stage 1: Determine the lightbulb situation for digit 1 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 1 in 2 ways Stage 2: Determine the lightbulb situation for digit 2 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 2 in 2 ways Stage 3: Determine the lightbulb situation for digit 3 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 3 in 2 ways . . . Stage 9: Determine the lightbulb situation for digit 9 There are only two possibilities here: the lightbulb is ON or it's OFF So, we can complete stage 9 in 2 ways By the Fundamental Counting Principle (FCP), we can complete all 9 stages (and thus create a number with its digits in ascending order) in (2)(2)(2)(2)(2)(2)(2)(2)(2) ways (= 512 ways IMPORTANT: Among the 512 possibilities, we have accidentally included some scenarios that don't meet the conditions in the question. For example, among the 512 possibilities, we have accidentally included 1 scenario where ALL of the lightbulbs are off. In this case, there is NO resulting number. Likewise, the question tells us that we must have 2 or more digits, and among the 512 possibilities, we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is off (resulting in 1, 2, 3, 4,...8, and 9). So, the number of numbers that meet the given conditions = 512 - 1 - 9 = 502 = C Cheers, Brent Aside: For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775 Hi Brent, Brilliant approach,as usual..but in the solution you suggested, are you meaning to say that "we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is 'ON' (resulting in 1, 2, 3, 4,...8, and 9)?? Thanks. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12677 messages Followed by: 1246 members Upvotes: 5254 GMAT Score: 770 parveen110 wrote: Hi Brent, Brilliant approach,as usual..but in the solution you suggested, are you meaning to say that "we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is 'ON' (resulting in 1, 2, 3, 4,...8, and 9)?? Thanks. Good catch, parveen110, thanks! I've edited my solution accordingly. Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! Legendary Member Joined 12 May 2015 Posted: 518 messages Upvotes: 10 Test Date: 3 Oct Target GMAT Score: 750 This question seems too tough for GMAT. Newbie | Next Rank: 10 Posts Joined 07 May 2017 Posted: 1 messages just wondering how to formulate that digit..i am seeing an explanation but can’t understand the ‘ask’ here. n >= 2 digits does it mean 2 digit or 3 or 4,5,6,7,..keeps going on. what is the limitation that we apply or how best can this be broken? why should we consider 123456789 and not any repetitions witin it. (looks like old post but just stuck with this understanding) ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12677 messages Followed by: 1246 members Upvotes: 5254 GMAT Score: 770 vinsmba wrote: just wondering how to formulate that digit..i am seeing an explanation but can’t understand the ‘ask’ here. n >= 2 digits does it mean 2 digit or 3 or 4,5,6,7,..keeps going on. what is the limitation that we apply or how best can this be broken? why should we consider 123456789 and not any repetitions witin it. (looks like old post but just stuck with this understanding) Can you rephrase your question please. Aside: "of two or more than two digits" means "two or more digits" (if that's what you're asking) Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12677 messages Followed by: 1246 members Upvotes: 5254 GMAT Score: 770 nikhilgmat31 wrote: This question seems too tough for GMAT. Quite possibly. If it is within the scope of the GMAT, it's a 750+ level question. Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use our video course along with Sign up for our free Question of the Day emails And check out all of our free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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