Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
Login or Register

5-digit positive integer

This topic has 11 member replies
goyalsau GMAT Destroyer!
Joined
02 Aug 2010
Posted:
866 messages
Thanked:
27 times
Target GMAT Score:
700+
5-digit positive integer Post Sun Oct 17, 2010 4:31 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    How many 5-digit positive integers exist where no two consecutive digits are the same ?

    9*9*8*7*6
    9*9*8*8*8
    10*9^^4
    9^^5
    9*8^^4

    _________________
    Saurabh Goyal
    talk_to_saurabh@yahoo.com
    -------------------------


    EveryBody Wants to Win But Nobody wants to prepare for Win.

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    shovan85 Community Manager
    Joined
    23 Sep 2010
    Posted:
    991 messages
    Followed by:
    23 members
    Thanked:
    142 times
    Post Sun Oct 17, 2010 4:46 am
    IMO D

    ABCDE be the number

    A can be selected in 9 ways [not 0 rest 1-9]
    B can be selected in 9 ways [except the selceted digit for A rest all including 0]
    C can be selected in 9 ways
    D can be selected in 9 ways
    E can be selected in 9 ways

    Total 9*9*9*9*9 = 9^5

    goyalsau GMAT Destroyer!
    Joined
    02 Aug 2010
    Posted:
    866 messages
    Thanked:
    27 times
    Target GMAT Score:
    700+
    Post Sun Oct 17, 2010 5:36 am
    shovan85 wrote:
    IMO D

    ABCDE be the number

    A can be selected in 9 ways [not 0 rest 1-9]
    B can be selected in 9 ways [except the selceted digit for A rest all including 0]
    C can be selected in 9 ways
    D can be selected in 9 ways
    E can be selected in 9 ways

    Total 9*9*9*9*9 = 9^5
    I would definitely like to recommend you to write CAT(You have all the answers ) ,
    I think you are 98+ percentile Guy.

    Take it as a complement........

    _________________
    Saurabh Goyal
    talk_to_saurabh@yahoo.com
    -------------------------


    EveryBody Wants to Win But Nobody wants to prepare for Win.

    Thanked by: shovan85

    GMAT/MBA Expert

    fskilnik GMAT Instructor
    Joined
    09 Oct 2010
    Posted:
    306 messages
    Followed by:
    21 members
    Thanked:
    54 times
    Post Sun Oct 17, 2010 5:33 pm
    Good solution indeed, shovan. Clear and to-the-point.

    Let us put some spice in the problem, shall we?

    How many 5-digit positive integers exist in which there are exactly two consecutive digits that are the same ?

    Regards,
    Fabio.

    Hint/Sugestion: there is the possibility of choosing the first two, then the second and third digits, etc... but there is a trick to "glue" two of the digits as if they were just one to apply shovan´s idea... think about it!)

    _________________
    GMATH high-level Quant Prep
    www.GMATH.NET

    shovan85 Community Manager
    Joined
    23 Sep 2010
    Posted:
    991 messages
    Followed by:
    23 members
    Thanked:
    142 times
    Post Mon Oct 18, 2010 2:31 am
    fskilnik wrote:
    Good solution indeed, shovan. Clear and to-the-point.

    Hi FSKILNIK,

    I have a concern about the actual problem. I could solve it by starting from A then B then C D E.
    ABCDE be the number

    A can be selected in 9 ways [not 0 rest 1-9]
    B can be selected in 9 ways [except the selceted digit for A rest all including 0]
    C can be selected in 9 ways
    D can be selected in 9 ways
    E can be selected in 9 ways

    Total 9*9*9*9*9 = 9^5

    But if I start from E (backwards) D C B A.... Its quite possible to solve but why the results do not match?
    I mean now E can be selected in 10 ways right (0-9)

    _________________
    If the problem is Easy Respect it, if the problem is tough Attack it

    GMAT/MBA Expert

    fskilnik GMAT Instructor
    Joined
    09 Oct 2010
    Posted:
    306 messages
    Followed by:
    21 members
    Thanked:
    54 times
    Post Mon Oct 18, 2010 3:22 am
    shovan85 wrote:
    Hi FSKILNIK,

    I have a concern about the actual problem. I could solve it by starting from A then B then C D E.

    But if I start from E (backwards) D C B A.... Its quite possible to solve but why the results do not match?
    I mean now E can be selected in 10 ways right (0-9)
    Good question, shovan85! Thank you for the opportunity you give me to try to clear that for you...

    One of the "golden rules" of Combinatorics is "satisfy the restrictions first", perhaps because it is hard (sometimes impossible) to manage the difficulties if you let them "propagate"...

    In this case, if you start with the first digit, the problem with 0 to be excluded from this (first) place is managed at first, what is according to the "golden rule", right?

    If you start with the last digit, you see the problem: at the moment you have to choose how many possibilities you have for A given that you have already reasoned for E, D, C and B... you have a nightmare:

    (1) If B was 0, then A will be not "naturally" (great) but...
    (2) If B was not 0, then A CANNOT be equal to B NOR to zero...

    You see, you would need to separate in 2 cases, because the Multiplicative Principle could not be applied as in the first way you chose to answer the problem!!

    I hope I did clear it all for you. If not, please ask me about my explanation or about what you still didn´t get, ok?!

    Best Regards and thank you for your trust!
    Fabio.

    _________________
    GMATH high-level Quant Prep
    www.GMATH.NET

    Thanked by: shovan85
    shovan85 Community Manager
    Joined
    23 Sep 2010
    Posted:
    991 messages
    Followed by:
    23 members
    Thanked:
    142 times
    Post Mon Oct 18, 2010 3:37 am
    fskilnik wrote:
    One of the "golden rules" of Combinatorics is "satisfy the restrictions first", perhaps because it is hard (sometimes impossible) to manage the difficulties if you let them "propagate"...

    In this case, if you start with the first digit, the problem with 0 to be excluded from this (first) place is managed at first, what is according to the "golden rule", right?

    So we have to restrict it first. I was thinking if I start from E to A

    Will it be like 10*9*9*9*9 - 9*9*9*9? I was subtracting all possible 4 digit numbers when A becomes zero. If you see now we can have the answer equal to 9^5.

    I know this discussion is out of context and in GMAT I ll not try to experiment. But what do you say for my above reasoning?
    If its correct then I ll gain more confidence on this as I can understand the problem very well.

    _________________
    If the problem is Easy Respect it, if the problem is tough Attack it

    shovan85 Community Manager
    Joined
    23 Sep 2010
    Posted:
    991 messages
    Followed by:
    23 members
    Thanked:
    142 times
    Post Mon Oct 18, 2010 4:00 am
    fskilnik wrote:
    How many 5-digit positive integers exist in which there are exactly two consecutive digits that are the same ?
    ABCDE now AB has to be equal then BC then CD then DE.

    AB can be selected in 9 ways (except zero) and C, D and E each can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    BC can be selected in 9 ways and A (except zero) D, and E each can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    CD can be selected in 9 ways and A (except zero) B, and E can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    DE can be selected in 9 ways and A (except zero) B, and C can be selected in 9 ways = 9*9*9*9 = 9^4 ways

    So total = 4*9^4

    _________________
    If the problem is Easy Respect it, if the problem is tough Attack it

    GMAT/MBA Expert

    fskilnik GMAT Instructor
    Joined
    09 Oct 2010
    Posted:
    306 messages
    Followed by:
    21 members
    Thanked:
    54 times
    Post Mon Oct 18, 2010 4:00 am
    shovan85 wrote:
    Will it be like 10*9*9*9*9 - 9*9*9*9? I was subtracting all possible 4 digit numbers when A becomes zero. If you see now we can have the answer equal to 9^5.
    You are right to think about alternative options, that´s the proper way of gaining experience and also of understanding why "good paths/strategies/techniques are really good" (because they avoid headaches that bad ones don´t...)

    About your calculations... they are right! Being more explicit:

    10*9*9*9*9 is 10 options for E, 9 for D, 9 for C, 9 for B (till now perfect) but when it comes to 9 for A, we know that we are making some mistakes, that we are counting also when A is zero, and that is not allowed... your idea is to take out exactly what was "extra-counted" (correct), and then you forced A to be zero (1 option for A), then 9 options for B, 9 for C, 9 for D and 9 for E... and that is PERFECT. Do you see it clearer now?

    Best Regards,
    Fábio.

    P.S.: you are right to check your reasoning even when the result is the same. I know same examples of wrong reasoning coming to right answer, LoL...

    _________________
    GMATH high-level Quant Prep
    www.GMATH.NET

    shovan85 Community Manager
    Joined
    23 Sep 2010
    Posted:
    991 messages
    Followed by:
    23 members
    Thanked:
    142 times
    Post Mon Oct 18, 2010 4:06 am
    fskilnik wrote:
    shovan85 wrote:
    Will it be like 10*9*9*9*9 - 9*9*9*9? I was subtracting all possible 4 digit numbers when A becomes zero. If you see now we can have the answer equal to 9^5.
    You are right to think about alternative options, that´s the proper way of gaining experience and also of understanding why "good paths/strategies/techniques are really good" (because they avoid headaches that bad ones don´t...)

    About your calculations... they are right! Being more explicit:

    10*9*9*9*9 is 10 options for E, 9 for D, 9 for C, 9 for B (till now perfect) but when it comes to 9 for A, we know that we are making some mistakes, that we are counting also when A is zero, and that is not allowed... your idea is to take out exactly what was "extra-counted" (correct), and then you forced A to be zero (1 option for A), then 9 options for B, 9 for C, 9 for D and 9 for E... and that is PERFECT. Do you see it clearer now?

    Best Regards,
    Fábio.

    P.S.: you are right to check your reasoning even when the result is the same. I know same examples of wrong reasoning coming to right answer, LoL...
    Thanks this really helps. As per wrong reasoning to right answer I always do that in Verbal Section Wink

    _________________
    If the problem is Easy Respect it, if the problem is tough Attack it

    GMAT/MBA Expert

    fskilnik GMAT Instructor
    Joined
    09 Oct 2010
    Posted:
    306 messages
    Followed by:
    21 members
    Thanked:
    54 times
    Post Mon Oct 18, 2010 4:13 am
    shovan85 wrote:
    fskilnik wrote:
    How many 5-digit positive integers exist in which there are exactly two consecutive digits that are the same ?
    ABCDE now AB has to be equal then BC then CD then DE.

    AB can be selected in 9 ways (except zero) and C, D and E each can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    BC can be selected in 9 ways and A (except zero) D, and E each can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    CD can be selected in 9 ways and A (except zero) B, and E can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    DE can be selected in 9 ways and A (except zero) B, and C can be selected in 9 ways = 9*9*9*9 = 9^4 ways

    So total = 4*9^4
    Nice! Just to emphasize the "restrictions first", I would write:

    shovan85 wrote:
    BC can be selected in 9 ways and A (except zero) D, and E each can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    As

    Quote:
    A selected in 9 ways (except zero), then BC, D, and E each can be selected in 9 ways = 9*9*9*9 = 9^4 ways
    but of course it´s just a matter of preference!

    What about my idea of gluing? If you consider (say) AB as a single block, you would have 4 blocks to be represented by different digits... try to finish the reasoning from now, shovan, but remember not to give the block "AB" special treatment! Wink

    _________________
    GMATH high-level Quant Prep
    www.GMATH.NET

    GMAT/MBA Expert

    fskilnik GMAT Instructor
    Joined
    09 Oct 2010
    Posted:
    306 messages
    Followed by:
    21 members
    Thanked:
    54 times
    Post Mon Oct 18, 2010 4:24 am
    shovan85 wrote:
    Thanks this really helps. As per wrong reasoning to right answer I always do that in Verbal Section Wink
    Great, so now add Math to it Wink

    P.S.: in teaching/learning Combinatorics, we have two types of mistakes:

    (1) We missed someone in the counting
    (2) We counted someone (at least) twice, what I like to call double-counting´s...

    That understood, WHENEVER you are not sure about your (potential) solution, try to understand if your solution avoids the mistakes mentioned above. If you are not sure, that means you should probably try another (easier/clearer) approach, in which you (will be able to) see, clearly, that you are not making mistakes of these types.

    _________________
    GMATH high-level Quant Prep
    www.GMATH.NET

    Best Conversation Starters

    1 abhasjha 52 topics
    2 tanvis1120 21 topics
    3 qwerty12321 20 topics
    4 Nijo 18 topics
    5 sapuna 13 topics
    See More Top Beat The GMAT Members...

    Most Active Experts

    1 image description GMATGuruNY

    The Princeton Review Teacher

    180 posts
    2 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    151 posts
    3 image description Jim@StratusPrep

    Stratus Prep

    60 posts
    4 image description David@GMATPrepNow

    GMAT Prep Now Teacher

    51 posts
    5 image description Jon@Admissionado

    Admissionado

    46 posts
    See More Top Beat The GMAT Experts