How many odd three-digits numbers are there whose all three digits are different

Multi tool use
Multi tool use












4












$begingroup$


I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19
















4












$begingroup$


I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19














4












4








4


2



$begingroup$


I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory







share|cite|improve this question









$endgroup$




I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory




combinatorics number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 9 '17 at 8:33









someone123123someone123123

428314




428314








  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19














  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19








1




1




$begingroup$
The mistake is you might have used 0 for the middle digit. So there are two cases.
$endgroup$
– quasi
Jul 9 '17 at 8:36






$begingroup$
The mistake is you might have used 0 for the middle digit. So there are two cases.
$endgroup$
– quasi
Jul 9 '17 at 8:36














$begingroup$
I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
$endgroup$
– someone123123
Jul 9 '17 at 8:39




$begingroup$
I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
$endgroup$
– someone123123
Jul 9 '17 at 8:39




2




2




$begingroup$
@someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
$endgroup$
– IanF1
Jul 9 '17 at 15:34




$begingroup$
@someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
$endgroup$
– IanF1
Jul 9 '17 at 15:34












$begingroup$
To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
$endgroup$
– Daniel Wagner
Jul 9 '17 at 23:19




$begingroup$
To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
$endgroup$
– Daniel Wagner
Jul 9 '17 at 23:19










4 Answers
4






active

oldest

votes


















13












$begingroup$

Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




  • $5$ choices for the last digit (to make the number odd)


  • $8$ choices left for the first digit (since $0$ can't be used)


  • $8$ choices still left for the second digit (since $0$ is now available for use)



Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






share|cite|improve this answer











$endgroup$





















    10












    $begingroup$

    The mistake is you might have used $0$ for the middle digit. So there are two cases.



    Case ($1$): Using $0$ for the middle digit:



    $qquad$The count is $5,{times},1,{times},8=40$.



    Case ($2$): Not using $0$ for the middle digit:



    $qquad$The count is $5,{times},8,{times},7=280$.



    So you get a total count of $40+280=320$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
      $endgroup$
      – someone123123
      Jul 9 '17 at 8:44






    • 6




      $begingroup$
      Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
      $endgroup$
      – quasi
      Jul 9 '17 at 8:46





















    1












    $begingroup$

    Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
    Case (1): Using 0 for the middle digit.



    The count is 5×1×8=40.



    Case (2): Not using 0 for the middle digit.



    The count is 5×8×7=280.



    So you get a total count of 40+280=320.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Welcome to MSE. Your answer adds nothing new to the already existing answers.
      $endgroup$
      – José Carlos Santos
      Jan 17 at 17:54



















    -1












    $begingroup$

    Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



    With repetition of digits we would have had 5 * 5 * 5 = 125



    But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



    And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




    5 * 4 * 3 = 60 is my answer






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
      $endgroup$
      – someone123123
      Jul 12 '17 at 10:13











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2352236%2fhow-many-odd-three-digits-numbers-are-there-whose-all-three-digits-are-different%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    13












    $begingroup$

    Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



    Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




    • $5$ choices for the last digit (to make the number odd)


    • $8$ choices left for the first digit (since $0$ can't be used)


    • $8$ choices still left for the second digit (since $0$ is now available for use)



    Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






    share|cite|improve this answer











    $endgroup$


















      13












      $begingroup$

      Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



      Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




      • $5$ choices for the last digit (to make the number odd)


      • $8$ choices left for the first digit (since $0$ can't be used)


      • $8$ choices still left for the second digit (since $0$ is now available for use)



      Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






      share|cite|improve this answer











      $endgroup$
















        13












        13








        13





        $begingroup$

        Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



        Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




        • $5$ choices for the last digit (to make the number odd)


        • $8$ choices left for the first digit (since $0$ can't be used)


        • $8$ choices still left for the second digit (since $0$ is now available for use)



        Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






        share|cite|improve this answer











        $endgroup$



        Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



        Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




        • $5$ choices for the last digit (to make the number odd)


        • $8$ choices left for the first digit (since $0$ can't be used)


        • $8$ choices still left for the second digit (since $0$ is now available for use)



        Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 9 '17 at 15:15

























        answered Jul 9 '17 at 9:16









        true blue aniltrue blue anil

        20.5k11941




        20.5k11941























            10












            $begingroup$

            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46


















            10












            $begingroup$

            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46
















            10












            10








            10





            $begingroup$

            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.






            share|cite|improve this answer











            $endgroup$



            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 9 '17 at 8:45

























            answered Jul 9 '17 at 8:41









            quasiquasi

            36k22663




            36k22663












            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46




















            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46


















            $begingroup$
            Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
            $endgroup$
            – someone123123
            Jul 9 '17 at 8:44




            $begingroup$
            Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
            $endgroup$
            – someone123123
            Jul 9 '17 at 8:44




            6




            6




            $begingroup$
            Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
            $endgroup$
            – quasi
            Jul 9 '17 at 8:46






            $begingroup$
            Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
            $endgroup$
            – quasi
            Jul 9 '17 at 8:46













            1












            $begingroup$

            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54
















            1












            $begingroup$

            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54














            1












            1








            1





            $begingroup$

            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.






            share|cite|improve this answer









            $endgroup$



            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 17 at 17:26









            AJ WARRIORAJ WARRIOR

            112




            112












            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54


















            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54
















            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Jan 17 at 17:54




            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Jan 17 at 17:54











            -1












            $begingroup$

            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13
















            -1












            $begingroup$

            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13














            -1












            -1








            -1





            $begingroup$

            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer






            share|cite|improve this answer









            $endgroup$



            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jul 11 '17 at 21:57









            korwalskiykorwalskiy

            1091




            1091








            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13














            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13








            1




            1




            $begingroup$
            I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
            $endgroup$
            – someone123123
            Jul 12 '17 at 10:13




            $begingroup$
            I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
            $endgroup$
            – someone123123
            Jul 12 '17 at 10:13


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2352236%2fhow-many-odd-three-digits-numbers-are-there-whose-all-three-digits-are-different%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            7ooZTIErH3ITQ,u2Flrp6RpkODt,oebCJojAaPz,wW,R4xcUvu46JospBfias7UClV,Oq,WdeIbeYVMyqvJcn dzjiUr,6aJ a830134sH4
            -
            00Ytqw8RQD,6O1nXPVEy i2DAimb,k,PJNR,ebBMhWbkh3ezdfN6impJVn gpiNQlrLvE7TFMxRPfmU

            Popular posts from this blog

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            menu

            Image
            menu    Wanted List (rus) Main Projects    FAMI Dumping project    Pirate cart's ID numbers list    My carts collection    Unpublished NES cheats    >All titles     Numbered titles     A titles     B titles     C titles     D titles     E titles     F titles     G titles     H titles     I titles     J titles     K titles     L titles     M titles     N titles     O titles     P titles     Q titles     R titles     S titles     T titles     U titles     V titles     W titles     X titles     Y titles     Z titles     Unique cheats     Buried secrets     Fixed generic cheats     Generic cheats     Secret passwords     Password generators     List of games with no cheats found    Unpublished Game Boy cheats    Unpublished Game Boy Color cheats Misc Stuff    Pirate Market    BONZA    FORTUNA    Decrypted SmyNES ROMs    NES inside (rus)    FCEUMM project Blog    cah4e3.wordpress.com Mail me:    cah4e3 at gmail dot c...
            Read more

            Dobbiaco

            Image
            Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
            Read more