Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?












3














Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?



one of the digits which can be formed is $4444$



$4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$



is there a well defined technique to solve this question.






combinatorics







share|cite|improve this question





























    3














    Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?



    one of the digits which can be formed is $4444$



    $4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$



    is there a well defined technique to solve this question.






    combinatorics







    share|cite|improve this question



























      3












      3








      3


      1





      Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?



      one of the digits which can be formed is $4444$



      $4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$



      is there a well defined technique to solve this question.






      combinatorics







      share|cite|improve this question















      Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?



      one of the digits which can be formed is $4444$



      $4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$



      is there a well defined technique to solve this question.






      combinatorics




      combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 26 '12 at 3:54

























      asked Aug 26 '12 at 3:40









      HOLYBIBLETHE

      1,45221837




      1,45221837






















          3 Answers
          3






          active

          oldest

          votes


















          1














          You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.



          First place can have either $3$ or $4$. So we have two choices. Lets analyse it.



          If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s



          So total = $3 times 3times 3 - 8 = 19$



          If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s



          So total = $3 times 3times 3 - 2 = 25$



          So, total choices = $19+25 = 44$, which is your answer.






          share|cite|improve this answer























          • There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
            – ladaghini
            Aug 26 '12 at 7:21












          • @ladaghini: please see the edited answer. I previously applied a wrong approach.
            – Rahul
            Aug 26 '12 at 7:54










          • But 44 is not an option. The correct answer is given as 51. I wonder how.
            – Ramit
            Dec 30 '16 at 4:53



















          1














          It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.






          share|cite|improve this answer





















          • Can you provide your final answer please?
            – Ramit
            Dec 30 '16 at 4:54










          • @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
            – Ross Millikan
            Dec 30 '16 at 5:29



















          0














          I will use the fundamental principle of counting to solve this question.



          Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.



          Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.



          We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.



          Case 1: Thousandth's place is 3.



          Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.



          Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).



          Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.



          Case 2: Thousandth's place is 4.



          Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.



          Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)



          Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.



          Summing each case up, we get $26 + 25 = 51$.



          Hope this helps.






          share|cite|improve this answer























            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%2f186968%2fgiven-digits-2-2-3-3-4-4-4-4-how-many-distinct-4-digit-numbers-greater-than%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.



            First place can have either $3$ or $4$. So we have two choices. Lets analyse it.



            If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s



            So total = $3 times 3times 3 - 8 = 19$



            If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s



            So total = $3 times 3times 3 - 2 = 25$



            So, total choices = $19+25 = 44$, which is your answer.






            share|cite|improve this answer























            • There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
              – ladaghini
              Aug 26 '12 at 7:21












            • @ladaghini: please see the edited answer. I previously applied a wrong approach.
              – Rahul
              Aug 26 '12 at 7:54










            • But 44 is not an option. The correct answer is given as 51. I wonder how.
              – Ramit
              Dec 30 '16 at 4:53
















            1














            You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.



            First place can have either $3$ or $4$. So we have two choices. Lets analyse it.



            If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s



            So total = $3 times 3times 3 - 8 = 19$



            If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s



            So total = $3 times 3times 3 - 2 = 25$



            So, total choices = $19+25 = 44$, which is your answer.






            share|cite|improve this answer























            • There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
              – ladaghini
              Aug 26 '12 at 7:21












            • @ladaghini: please see the edited answer. I previously applied a wrong approach.
              – Rahul
              Aug 26 '12 at 7:54










            • But 44 is not an option. The correct answer is given as 51. I wonder how.
              – Ramit
              Dec 30 '16 at 4:53














            1












            1








            1






            You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.



            First place can have either $3$ or $4$. So we have two choices. Lets analyse it.



            If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s



            So total = $3 times 3times 3 - 8 = 19$



            If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s



            So total = $3 times 3times 3 - 2 = 25$



            So, total choices = $19+25 = 44$, which is your answer.






            share|cite|improve this answer














            You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.



            First place can have either $3$ or $4$. So we have two choices. Lets analyse it.



            If first place is $3$ - so we have to choose 3 digits from $(2,2,3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333,332,323,233,334,343,433)$ is not possible, because we have just one $3$'s and two $2$'s



            So total = $3 times 3times 3 - 8 = 19$



            If first place is $4$ - so now we have to choose 3 digits from $(2,2,3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s



            So total = $3 times 3times 3 - 2 = 25$



            So, total choices = $19+25 = 44$, which is your answer.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 5 '13 at 7:00









            learner

            3,37032167




            3,37032167










            answered Aug 26 '12 at 7:16









            Rahul

            399310




            399310












            • There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
              – ladaghini
              Aug 26 '12 at 7:21












            • @ladaghini: please see the edited answer. I previously applied a wrong approach.
              – Rahul
              Aug 26 '12 at 7:54










            • But 44 is not an option. The correct answer is given as 51. I wonder how.
              – Ramit
              Dec 30 '16 at 4:53


















            • There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
              – ladaghini
              Aug 26 '12 at 7:21












            • @ladaghini: please see the edited answer. I previously applied a wrong approach.
              – Rahul
              Aug 26 '12 at 7:54










            • But 44 is not an option. The correct answer is given as 51. I wonder how.
              – Ramit
              Dec 30 '16 at 4:53
















            There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
            – ladaghini
            Aug 26 '12 at 7:21






            There's a total of $binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high.
            – ladaghini
            Aug 26 '12 at 7:21














            @ladaghini: please see the edited answer. I previously applied a wrong approach.
            – Rahul
            Aug 26 '12 at 7:54




            @ladaghini: please see the edited answer. I previously applied a wrong approach.
            – Rahul
            Aug 26 '12 at 7:54












            But 44 is not an option. The correct answer is given as 51. I wonder how.
            – Ramit
            Dec 30 '16 at 4:53




            But 44 is not an option. The correct answer is given as 51. I wonder how.
            – Ramit
            Dec 30 '16 at 4:53











            1














            It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.






            share|cite|improve this answer





















            • Can you provide your final answer please?
              – Ramit
              Dec 30 '16 at 4:54










            • @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
              – Ross Millikan
              Dec 30 '16 at 5:29
















            1














            It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.






            share|cite|improve this answer





















            • Can you provide your final answer please?
              – Ramit
              Dec 30 '16 at 4:54










            • @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
              – Ross Millikan
              Dec 30 '16 at 5:29














            1












            1








            1






            It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.






            share|cite|improve this answer












            It is a matter of detailed counting. You have to start with a $3$ or $4$ to be greater than $3000$. If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$. If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 26 '12 at 4:09









            Ross Millikan

            291k23196370




            291k23196370












            • Can you provide your final answer please?
              – Ramit
              Dec 30 '16 at 4:54










            • @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
              – Ross Millikan
              Dec 30 '16 at 5:29


















            • Can you provide your final answer please?
              – Ramit
              Dec 30 '16 at 4:54










            • @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
              – Ross Millikan
              Dec 30 '16 at 5:29
















            Can you provide your final answer please?
            – Ramit
            Dec 30 '16 at 4:54




            Can you provide your final answer please?
            – Ramit
            Dec 30 '16 at 4:54












            @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
            – Ross Millikan
            Dec 30 '16 at 5:29




            @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$?
            – Ross Millikan
            Dec 30 '16 at 5:29











            0














            I will use the fundamental principle of counting to solve this question.



            Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.



            Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.



            We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.



            Case 1: Thousandth's place is 3.



            Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.



            Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).



            Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.



            Case 2: Thousandth's place is 4.



            Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.



            Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)



            Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.



            Summing each case up, we get $26 + 25 = 51$.



            Hope this helps.






            share|cite|improve this answer




























              0














              I will use the fundamental principle of counting to solve this question.



              Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.



              Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.



              We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.



              Case 1: Thousandth's place is 3.



              Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.



              Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).



              Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.



              Case 2: Thousandth's place is 4.



              Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.



              Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)



              Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.



              Summing each case up, we get $26 + 25 = 51$.



              Hope this helps.






              share|cite|improve this answer


























                0












                0








                0






                I will use the fundamental principle of counting to solve this question.



                Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.



                Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.



                We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.



                Case 1: Thousandth's place is 3.



                Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.



                Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).



                Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.



                Case 2: Thousandth's place is 4.



                Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.



                Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)



                Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.



                Summing each case up, we get $26 + 25 = 51$.



                Hope this helps.






                share|cite|improve this answer














                I will use the fundamental principle of counting to solve this question.



                Given the set of numbers, $D = {2, 2, 3, 3, 3, 4, 4, 4, 4}$.



                Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.



                We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.



                Case 1: Thousandth's place is 3.



                Set, $D' = {2, 2, 3, 3, 4, 4, 4, 4}$.



                Rest of the 3 places can be filled in $3 times 3 times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).



                Therefore, ways to fill = $3 times 3 times 3 - (1+1) = 25$.



                Case 2: Thousandth's place is 4.



                Set, $D' = {2, 2, 3, 3, 3, 4, 4, 4}$.



                Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)



                Therefore ways to fill = $3 times 3 times 3 - 1 = 26$.



                Summing each case up, we get $26 + 25 = 51$.



                Hope this helps.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 7 at 13:04

























                answered Oct 26 at 4:19









                Manoj Baishya

                12




                12






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f186968%2fgiven-digits-2-2-3-3-4-4-4-4-how-many-distinct-4-digit-numbers-greater-than%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







                    -

                    Popular posts from this blog

                    Bressuire

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more

                    Cabo Verde

                    Image
                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                    Read more

                    Gyllenstierna

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more