Algebraic proof of $sum_{i=0}^k{{n choose i}{m choose {k-i}}}= {{m+n}choose k}$












2












$begingroup$


I can't figure out an algebraic proof for the following identity:



$$sum_{i=0}^k{{n choose i}{m choose {k-i}}}= {{m+n}choose k}$$



Combinatorical solution:



We can see that as choosing some from $n$ and the rest of $k$ from $m$, thus $k$ in total.



Or we could just choose $k$ from the union.






combinatorics discrete-mathematics summation binomial-coefficients







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    For an algebraic proof (I don't know why you want one, to be honest), but consider the coefficient of $x^k$ in $(1+x)^n(1+x)^m = (1+x)^{n+m}$.
    $endgroup$
    – Andrew D
    Sep 11 '13 at 17:11










  • $begingroup$
    Its called the Chu-Vandermonde Identity, and there's already a post about it here: math.stackexchange.com/questions/219928/…
    $endgroup$
    – Salieri
    Nov 23 '13 at 13:48






  • 1




    $begingroup$
    @AndrewD: The algebraic proof does not depend on $n,m$ being natural numbers (but the binomial powers are now formal powers series); this is one reason why one might want an algebraic proof.
    $endgroup$
    – Marc van Leeuwen
    Nov 23 '13 at 14:57
















2












$begingroup$


I can't figure out an algebraic proof for the following identity:



$$sum_{i=0}^k{{n choose i}{m choose {k-i}}}= {{m+n}choose k}$$



Combinatorical solution:



We can see that as choosing some from $n$ and the rest of $k$ from $m$, thus $k$ in total.



Or we could just choose $k$ from the union.






combinatorics discrete-mathematics summation binomial-coefficients







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    For an algebraic proof (I don't know why you want one, to be honest), but consider the coefficient of $x^k$ in $(1+x)^n(1+x)^m = (1+x)^{n+m}$.
    $endgroup$
    – Andrew D
    Sep 11 '13 at 17:11










  • $begingroup$
    Its called the Chu-Vandermonde Identity, and there's already a post about it here: math.stackexchange.com/questions/219928/…
    $endgroup$
    – Salieri
    Nov 23 '13 at 13:48






  • 1




    $begingroup$
    @AndrewD: The algebraic proof does not depend on $n,m$ being natural numbers (but the binomial powers are now formal powers series); this is one reason why one might want an algebraic proof.
    $endgroup$
    – Marc van Leeuwen
    Nov 23 '13 at 14:57














2












2








2


2



$begingroup$


I can't figure out an algebraic proof for the following identity:



$$sum_{i=0}^k{{n choose i}{m choose {k-i}}}= {{m+n}choose k}$$



Combinatorical solution:



We can see that as choosing some from $n$ and the rest of $k$ from $m$, thus $k$ in total.



Or we could just choose $k$ from the union.






combinatorics discrete-mathematics summation binomial-coefficients







share|cite|improve this question











$endgroup$




I can't figure out an algebraic proof for the following identity:



$$sum_{i=0}^k{{n choose i}{m choose {k-i}}}= {{m+n}choose k}$$



Combinatorical solution:



We can see that as choosing some from $n$ and the rest of $k$ from $m$, thus $k$ in total.



Or we could just choose $k$ from the union.






combinatorics discrete-mathematics summation binomial-coefficients




combinatorics discrete-mathematics summation binomial-coefficients






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 3:02









Martin Sleziak

45k10122277




45k10122277










asked Sep 11 '13 at 17:08









NightRaNightRa

9991720




9991720








  • 6




    $begingroup$
    For an algebraic proof (I don't know why you want one, to be honest), but consider the coefficient of $x^k$ in $(1+x)^n(1+x)^m = (1+x)^{n+m}$.
    $endgroup$
    – Andrew D
    Sep 11 '13 at 17:11










  • $begingroup$
    Its called the Chu-Vandermonde Identity, and there's already a post about it here: math.stackexchange.com/questions/219928/…
    $endgroup$
    – Salieri
    Nov 23 '13 at 13:48






  • 1




    $begingroup$
    @AndrewD: The algebraic proof does not depend on $n,m$ being natural numbers (but the binomial powers are now formal powers series); this is one reason why one might want an algebraic proof.
    $endgroup$
    – Marc van Leeuwen
    Nov 23 '13 at 14:57














  • 6




    $begingroup$
    For an algebraic proof (I don't know why you want one, to be honest), but consider the coefficient of $x^k$ in $(1+x)^n(1+x)^m = (1+x)^{n+m}$.
    $endgroup$
    – Andrew D
    Sep 11 '13 at 17:11










  • $begingroup$
    Its called the Chu-Vandermonde Identity, and there's already a post about it here: math.stackexchange.com/questions/219928/…
    $endgroup$
    – Salieri
    Nov 23 '13 at 13:48






  • 1




    $begingroup$
    @AndrewD: The algebraic proof does not depend on $n,m$ being natural numbers (but the binomial powers are now formal powers series); this is one reason why one might want an algebraic proof.
    $endgroup$
    – Marc van Leeuwen
    Nov 23 '13 at 14:57








6




6




$begingroup$
For an algebraic proof (I don't know why you want one, to be honest), but consider the coefficient of $x^k$ in $(1+x)^n(1+x)^m = (1+x)^{n+m}$.
$endgroup$
– Andrew D
Sep 11 '13 at 17:11




$begingroup$
For an algebraic proof (I don't know why you want one, to be honest), but consider the coefficient of $x^k$ in $(1+x)^n(1+x)^m = (1+x)^{n+m}$.
$endgroup$
– Andrew D
Sep 11 '13 at 17:11












$begingroup$
Its called the Chu-Vandermonde Identity, and there's already a post about it here: math.stackexchange.com/questions/219928/…
$endgroup$
– Salieri
Nov 23 '13 at 13:48




$begingroup$
Its called the Chu-Vandermonde Identity, and there's already a post about it here: math.stackexchange.com/questions/219928/…
$endgroup$
– Salieri
Nov 23 '13 at 13:48




1




1




$begingroup$
@AndrewD: The algebraic proof does not depend on $n,m$ being natural numbers (but the binomial powers are now formal powers series); this is one reason why one might want an algebraic proof.
$endgroup$
– Marc van Leeuwen
Nov 23 '13 at 14:57




$begingroup$
@AndrewD: The algebraic proof does not depend on $n,m$ being natural numbers (but the binomial powers are now formal powers series); this is one reason why one might want an algebraic proof.
$endgroup$
– Marc van Leeuwen
Nov 23 '13 at 14:57










3 Answers
3






active

oldest

votes


















1












$begingroup$

For identities involving binomial coefficients sometimes "combinatorial proofs" and "algebraic proofs" are offered. Here we have two proofs for the Vandermonde Convolution:




  1. Algebraic proof: The sum of the LHS of the convolution formula equals the coefficient of $x^k$ on the LHS of the polynomial
    $$
    (1+x)^n(1+x)^m=(1+x)^{m+n}.
    $$
    The binomial coefficient on the RHS of the convolutions formula equals the coefficient of $x^k$ on the RHS of the polynomial equation.


  2. Combinatorial proof: Suppose that there are $n+m$ objects in a set, $n$ of them white and $m$ of them black. There are $binom{n+m}{k}$ ways to choose $k$ elements in all. This is the RHS. The number of ways to choose $j$ white and $k-j$ black objects is the product $binom{n}{j}binom{m}{k-j}$, so the sum of all these objects, the LHS, must be the same total as on the RHS.







share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Using the binomial theorem consider $$sum_{k=0}^{m+n}{m+nchoose k}x^k=(1+x)^{m+n}.$$ The right-hand side becomes $$(1+x)^m(1+x)^n=(sum_{r=0}^m{mchoose r}x^r)(sum_{s=0}^n{nchoose s}x^s).$$ The product of the two polynomials on the right-hand side can be rewritten as $$sum_{k=0}^{m+n}(sum_{i=0}^k{nchoose i}{mchoose k-i})x^r.$$ Thus $$sum_{i=0}^k{nchoose i}{mchoose k-i}={m+nchoose k}.$$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      For another proof, see the First Proof of Theorem 3.29 in my Notes on the combinatorial fundamentals of algebra. (NB: It is Theorem 3.29 in the version of 10 January 2019. Theorem labels are subject to change, so search for "Chu-Vandermonde identity" if Theorem 3.29 is something different in your version.) This is not a particularly enlightening proof, but it has the advantage of requiring the least about $n$ and $m$: it works when $n$ and $m$ are elements of a commutative $mathbb{Q}$-algebra. (Compare to the combinatorial proof, which requires $n$ and $m$ to be nonnegative integers. Compare also to the proof using the binomial formula, which works generally only if one has an algebraic way of proving $left(1+xright)^nleft(1+xright)^m = left(1+xright)^{n+m}$ for non-integer $n$ and $m$.)



      This has been linked from How do i prove that $sumlimits_{r=0}^k binom{m}{r}binom{n}{k-r} = binom{m+n}{k}$ , which has other proofs.






      share|cite|improve this answer











      $endgroup$














        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%2f490741%2falgebraic-proof-of-sum-i-0kn-choose-im-choose-k-i-mn-choose%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












        $begingroup$

        For identities involving binomial coefficients sometimes "combinatorial proofs" and "algebraic proofs" are offered. Here we have two proofs for the Vandermonde Convolution:




        1. Algebraic proof: The sum of the LHS of the convolution formula equals the coefficient of $x^k$ on the LHS of the polynomial
          $$
          (1+x)^n(1+x)^m=(1+x)^{m+n}.
          $$
          The binomial coefficient on the RHS of the convolutions formula equals the coefficient of $x^k$ on the RHS of the polynomial equation.


        2. Combinatorial proof: Suppose that there are $n+m$ objects in a set, $n$ of them white and $m$ of them black. There are $binom{n+m}{k}$ ways to choose $k$ elements in all. This is the RHS. The number of ways to choose $j$ white and $k-j$ black objects is the product $binom{n}{j}binom{m}{k-j}$, so the sum of all these objects, the LHS, must be the same total as on the RHS.







        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          For identities involving binomial coefficients sometimes "combinatorial proofs" and "algebraic proofs" are offered. Here we have two proofs for the Vandermonde Convolution:




          1. Algebraic proof: The sum of the LHS of the convolution formula equals the coefficient of $x^k$ on the LHS of the polynomial
            $$
            (1+x)^n(1+x)^m=(1+x)^{m+n}.
            $$
            The binomial coefficient on the RHS of the convolutions formula equals the coefficient of $x^k$ on the RHS of the polynomial equation.


          2. Combinatorial proof: Suppose that there are $n+m$ objects in a set, $n$ of them white and $m$ of them black. There are $binom{n+m}{k}$ ways to choose $k$ elements in all. This is the RHS. The number of ways to choose $j$ white and $k-j$ black objects is the product $binom{n}{j}binom{m}{k-j}$, so the sum of all these objects, the LHS, must be the same total as on the RHS.







          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            For identities involving binomial coefficients sometimes "combinatorial proofs" and "algebraic proofs" are offered. Here we have two proofs for the Vandermonde Convolution:




            1. Algebraic proof: The sum of the LHS of the convolution formula equals the coefficient of $x^k$ on the LHS of the polynomial
              $$
              (1+x)^n(1+x)^m=(1+x)^{m+n}.
              $$
              The binomial coefficient on the RHS of the convolutions formula equals the coefficient of $x^k$ on the RHS of the polynomial equation.


            2. Combinatorial proof: Suppose that there are $n+m$ objects in a set, $n$ of them white and $m$ of them black. There are $binom{n+m}{k}$ ways to choose $k$ elements in all. This is the RHS. The number of ways to choose $j$ white and $k-j$ black objects is the product $binom{n}{j}binom{m}{k-j}$, so the sum of all these objects, the LHS, must be the same total as on the RHS.







            share|cite|improve this answer









            $endgroup$



            For identities involving binomial coefficients sometimes "combinatorial proofs" and "algebraic proofs" are offered. Here we have two proofs for the Vandermonde Convolution:




            1. Algebraic proof: The sum of the LHS of the convolution formula equals the coefficient of $x^k$ on the LHS of the polynomial
              $$
              (1+x)^n(1+x)^m=(1+x)^{m+n}.
              $$
              The binomial coefficient on the RHS of the convolutions formula equals the coefficient of $x^k$ on the RHS of the polynomial equation.


            2. Combinatorial proof: Suppose that there are $n+m$ objects in a set, $n$ of them white and $m$ of them black. There are $binom{n+m}{k}$ ways to choose $k$ elements in all. This is the RHS. The number of ways to choose $j$ white and $k-j$ black objects is the product $binom{n}{j}binom{m}{k-j}$, so the sum of all these objects, the LHS, must be the same total as on the RHS.








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 11 '13 at 18:39









            Dietrich BurdeDietrich Burde

            81.7k648106




            81.7k648106























                1












                $begingroup$

                Using the binomial theorem consider $$sum_{k=0}^{m+n}{m+nchoose k}x^k=(1+x)^{m+n}.$$ The right-hand side becomes $$(1+x)^m(1+x)^n=(sum_{r=0}^m{mchoose r}x^r)(sum_{s=0}^n{nchoose s}x^s).$$ The product of the two polynomials on the right-hand side can be rewritten as $$sum_{k=0}^{m+n}(sum_{i=0}^k{nchoose i}{mchoose k-i})x^r.$$ Thus $$sum_{i=0}^k{nchoose i}{mchoose k-i}={m+nchoose k}.$$






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  Using the binomial theorem consider $$sum_{k=0}^{m+n}{m+nchoose k}x^k=(1+x)^{m+n}.$$ The right-hand side becomes $$(1+x)^m(1+x)^n=(sum_{r=0}^m{mchoose r}x^r)(sum_{s=0}^n{nchoose s}x^s).$$ The product of the two polynomials on the right-hand side can be rewritten as $$sum_{k=0}^{m+n}(sum_{i=0}^k{nchoose i}{mchoose k-i})x^r.$$ Thus $$sum_{i=0}^k{nchoose i}{mchoose k-i}={m+nchoose k}.$$






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Using the binomial theorem consider $$sum_{k=0}^{m+n}{m+nchoose k}x^k=(1+x)^{m+n}.$$ The right-hand side becomes $$(1+x)^m(1+x)^n=(sum_{r=0}^m{mchoose r}x^r)(sum_{s=0}^n{nchoose s}x^s).$$ The product of the two polynomials on the right-hand side can be rewritten as $$sum_{k=0}^{m+n}(sum_{i=0}^k{nchoose i}{mchoose k-i})x^r.$$ Thus $$sum_{i=0}^k{nchoose i}{mchoose k-i}={m+nchoose k}.$$






                    share|cite|improve this answer











                    $endgroup$



                    Using the binomial theorem consider $$sum_{k=0}^{m+n}{m+nchoose k}x^k=(1+x)^{m+n}.$$ The right-hand side becomes $$(1+x)^m(1+x)^n=(sum_{r=0}^m{mchoose r}x^r)(sum_{s=0}^n{nchoose s}x^s).$$ The product of the two polynomials on the right-hand side can be rewritten as $$sum_{k=0}^{m+n}(sum_{i=0}^k{nchoose i}{mchoose k-i})x^r.$$ Thus $$sum_{i=0}^k{nchoose i}{mchoose k-i}={m+nchoose k}.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 26 '13 at 22:40

























                    answered Nov 23 '13 at 14:03









                    1233dfv1233dfv

                    4,1931326




                    4,1931326























                        0












                        $begingroup$

                        For another proof, see the First Proof of Theorem 3.29 in my Notes on the combinatorial fundamentals of algebra. (NB: It is Theorem 3.29 in the version of 10 January 2019. Theorem labels are subject to change, so search for "Chu-Vandermonde identity" if Theorem 3.29 is something different in your version.) This is not a particularly enlightening proof, but it has the advantage of requiring the least about $n$ and $m$: it works when $n$ and $m$ are elements of a commutative $mathbb{Q}$-algebra. (Compare to the combinatorial proof, which requires $n$ and $m$ to be nonnegative integers. Compare also to the proof using the binomial formula, which works generally only if one has an algebraic way of proving $left(1+xright)^nleft(1+xright)^m = left(1+xright)^{n+m}$ for non-integer $n$ and $m$.)



                        This has been linked from How do i prove that $sumlimits_{r=0}^k binom{m}{r}binom{n}{k-r} = binom{m+n}{k}$ , which has other proofs.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          For another proof, see the First Proof of Theorem 3.29 in my Notes on the combinatorial fundamentals of algebra. (NB: It is Theorem 3.29 in the version of 10 January 2019. Theorem labels are subject to change, so search for "Chu-Vandermonde identity" if Theorem 3.29 is something different in your version.) This is not a particularly enlightening proof, but it has the advantage of requiring the least about $n$ and $m$: it works when $n$ and $m$ are elements of a commutative $mathbb{Q}$-algebra. (Compare to the combinatorial proof, which requires $n$ and $m$ to be nonnegative integers. Compare also to the proof using the binomial formula, which works generally only if one has an algebraic way of proving $left(1+xright)^nleft(1+xright)^m = left(1+xright)^{n+m}$ for non-integer $n$ and $m$.)



                          This has been linked from How do i prove that $sumlimits_{r=0}^k binom{m}{r}binom{n}{k-r} = binom{m+n}{k}$ , which has other proofs.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            For another proof, see the First Proof of Theorem 3.29 in my Notes on the combinatorial fundamentals of algebra. (NB: It is Theorem 3.29 in the version of 10 January 2019. Theorem labels are subject to change, so search for "Chu-Vandermonde identity" if Theorem 3.29 is something different in your version.) This is not a particularly enlightening proof, but it has the advantage of requiring the least about $n$ and $m$: it works when $n$ and $m$ are elements of a commutative $mathbb{Q}$-algebra. (Compare to the combinatorial proof, which requires $n$ and $m$ to be nonnegative integers. Compare also to the proof using the binomial formula, which works generally only if one has an algebraic way of proving $left(1+xright)^nleft(1+xright)^m = left(1+xright)^{n+m}$ for non-integer $n$ and $m$.)



                            This has been linked from How do i prove that $sumlimits_{r=0}^k binom{m}{r}binom{n}{k-r} = binom{m+n}{k}$ , which has other proofs.






                            share|cite|improve this answer











                            $endgroup$



                            For another proof, see the First Proof of Theorem 3.29 in my Notes on the combinatorial fundamentals of algebra. (NB: It is Theorem 3.29 in the version of 10 January 2019. Theorem labels are subject to change, so search for "Chu-Vandermonde identity" if Theorem 3.29 is something different in your version.) This is not a particularly enlightening proof, but it has the advantage of requiring the least about $n$ and $m$: it works when $n$ and $m$ are elements of a commutative $mathbb{Q}$-algebra. (Compare to the combinatorial proof, which requires $n$ and $m$ to be nonnegative integers. Compare also to the proof using the binomial formula, which works generally only if one has an algebraic way of proving $left(1+xright)^nleft(1+xright)^m = left(1+xright)^{n+m}$ for non-integer $n$ and $m$.)



                            This has been linked from How do i prove that $sumlimits_{r=0}^k binom{m}{r}binom{n}{k-r} = binom{m+n}{k}$ , which has other proofs.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 10 at 1:49

























                            answered Jul 24 '15 at 20:14









                            darij grinbergdarij grinberg

                            11.5k33168




                            11.5k33168






























                                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%2f490741%2falgebraic-proof-of-sum-i-0kn-choose-im-choose-k-i-mn-choose%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