Find the number of ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy












5












$begingroup$


The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



$$square g square g square g square g square g square gsquare$$
Therefore, we come up with:



$$6!cdot 7P5=1space 814space 400$$



However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



$$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
$$5!cdot6P5cdot12=1space036space800$$



Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
$$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



$$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
In which we have:
$$2(5!)^2cdot6=172space800$$



So which is right? Is there anything wrong with each logic?



Thank you.






combinatorics permutations







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



    $$square g square g square g square g square g square gsquare$$
    Therefore, we come up with:



    $$6!cdot 7P5=1space 814space 400$$



    However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



    $$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



    We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
    $$5!cdot6P5cdot12=1space036space800$$



    Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
    $$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



    Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



    $$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
    In which we have:
    $$2(5!)^2cdot6=172space800$$



    So which is right? Is there anything wrong with each logic?



    Thank you.






    combinatorics permutations







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



      $$square g square g square g square g square g square gsquare$$
      Therefore, we come up with:



      $$6!cdot 7P5=1space 814space 400$$



      However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



      $$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



      We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
      $$5!cdot6P5cdot12=1space036space800$$



      Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
      $$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



      Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



      $$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
      In which we have:
      $$2(5!)^2cdot6=172space800$$



      So which is right? Is there anything wrong with each logic?



      Thank you.






      combinatorics permutations







      share|cite|improve this question











      $endgroup$




      The most straight-forward answer is to arrange 6 girls and put 5 boys in the remaining seven slots by order:



      $$square g square g square g square g square g square gsquare$$
      Therefore, we come up with:



      $$6!cdot 7P5=1space 814space 400$$



      However, I notice that if you take one of the girls away (let's label her as $^*$), we have:



      $$^*square^*g^*square^*g^*square^*g^*square^*g^*square^*g^*square^*$$



      We see that no matter where girl$^*$ stands, the boys will still not stand next to each others. Now we can arrange 5 girls and put all the boys in and put girl$^*$ in the 12 places by order:
      $$5!cdot6P5cdot12=1space036space800$$



      Let's say we have 5 boys and 5 girls, we then have two possible arrangements:
      $$gspace bspace gspace bspace gspace bspace gspace bspace gspace b + bspace gspace bspace gspace bspace gspace bspace gspace bspace g$$



      Now, we add in girl$^*$. Here we have 2 different ways of arranging 5 boys and 5 girls and we are left with six slots for girl$^*$ in each case:



      $$^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*b ^*+ ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*gspace ^*bspace ^*g ^*$$
      In which we have:
      $$2(5!)^2cdot6=172space800$$



      So which is right? Is there anything wrong with each logic?



      Thank you.






      combinatorics permutations




      combinatorics permutations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 48 mins ago







      Huy Tran Van

















      asked 1 hour ago









      Huy Tran VanHuy Tran Van

      264




      264






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Your first computation is correct.



          The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



          Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
            $endgroup$
            – Huy Tran Van
            22 mins ago












          • $begingroup$
            @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
            $endgroup$
            – orlp
            11 mins ago












          • $begingroup$
            I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
            $endgroup$
            – Huy Tran Van
            4 mins ago





















          1












          $begingroup$

          In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



          In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



          The right one is the first.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

            A $7 choose 5$ is calculated as
            $$
            frac{7!}{5!*(7-5)!} = 21
            $$
            Therefore there are 21 combinations in which the boys could be arranged.



            I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



            Picture of how the boys and girls could be arranged in 21 combinations






            share|cite|improve this answer








            New contributor




            Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
              $endgroup$
              – Tucktuckgoose
              35 mins ago











            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%2f3084008%2ffind-the-number-of-ways-that-5-boys-and-6-girls-can-stand-in-a-row-if-no-boy-sta%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









            2












            $begingroup$

            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              22 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              11 mins ago












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              4 mins ago


















            2












            $begingroup$

            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              22 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              11 mins ago












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              4 mins ago
















            2












            2








            2





            $begingroup$

            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.






            share|cite|improve this answer









            $endgroup$



            Your first computation is correct.



            The latter two computations undercount because they start off by taking one of the girls away. Let's say they always take girl #1 away.



            Now if we number the girls, those solutions can contain as an example $1 2 b 3 b 4 b 5 b 6b$. But they can not contain $1b2b34b5b6b$. We've lost the ability to have two girls next to eachother without involving girl #1.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 47 mins ago









            orlporlp

            7,4111330




            7,4111330












            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              22 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              11 mins ago












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              4 mins ago




















            • $begingroup$
              Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
              $endgroup$
              – Huy Tran Van
              22 mins ago












            • $begingroup$
              @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
              $endgroup$
              – orlp
              11 mins ago












            • $begingroup$
              I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
              $endgroup$
              – Huy Tran Van
              4 mins ago


















            $begingroup$
            Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
            $endgroup$
            – Huy Tran Van
            22 mins ago






            $begingroup$
            Now, I know the last two undercount somehow but I can't see why '$1b2b3b4b5b6b$' is not possible in the second case?
            $endgroup$
            – Huy Tran Van
            22 mins ago














            $begingroup$
            @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
            $endgroup$
            – orlp
            11 mins ago






            $begingroup$
            @HuyTranVan You misread. I said $1b2b34b5b6b$. There is no $b$ between $3$ and $4$.
            $endgroup$
            – orlp
            11 mins ago














            $begingroup$
            I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
            $endgroup$
            – Huy Tran Van
            4 mins ago






            $begingroup$
            I wanted to know why $12b3b4b5b6b$ is possible but then $1b2b34b5b6b$ (no b between 3 and 4) is not counted in the second case. According to you when one of the girls stands next to another, the question changed?
            $endgroup$
            – Huy Tran Van
            4 mins ago













            1












            $begingroup$

            In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



            In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



            The right one is the first.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



              In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



              The right one is the first.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



                In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



                The right one is the first.






                share|cite|improve this answer











                $endgroup$



                In the last one you're obviously leaving some out, which could be gotten by swapping some of bg combinations in either sum.



                In the second, also, you're not counting certain arrangements, which could be gotten by, again, swapping at least some bg combinations.



                The right one is the first.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 40 mins ago

























                answered 47 mins ago









                Chris CusterChris Custer

                11.5k3824




                11.5k3824























                    0












                    $begingroup$

                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations






                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      35 mins ago
















                    0












                    $begingroup$

                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations






                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      35 mins ago














                    0












                    0








                    0





                    $begingroup$

                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations






                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    You are on the right track with thinking that there are 7 possible slots with 5 boys to be arranged into those slots. This is a combination problem with can be treated as a 7 choose 5 or a $7 choose 5$.

                    A $7 choose 5$ is calculated as
                    $$
                    frac{7!}{5!*(7-5)!} = 21
                    $$
                    Therefore there are 21 combinations in which the boys could be arranged.



                    I have also used an excel spreadsheet to show the possible ways that they could be arranged which ends up equaling 21.



                    Picture of how the boys and girls could be arranged in 21 combinations







                    share|cite|improve this answer








                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 38 mins ago









                    TucktuckgooseTucktuckgoose

                    11




                    11




                    New contributor




                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Tucktuckgoose is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.












                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      35 mins ago


















                    • $begingroup$
                      My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                      $endgroup$
                      – Tucktuckgoose
                      35 mins ago
















                    $begingroup$
                    My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                    $endgroup$
                    – Tucktuckgoose
                    35 mins ago




                    $begingroup$
                    My answer is predicated that this is a combination problem, not a permutation problem. As they are described as boys and girls, they are likely interchangeable and this is a combination. If the question was how many ways can Tom, Dick, Harry, Sam, and Bob be arranged, that would be a different question with a much larger answer.
                    $endgroup$
                    – Tucktuckgoose
                    35 mins ago


















                    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%2f3084008%2ffind-the-number-of-ways-that-5-boys-and-6-girls-can-stand-in-a-row-if-no-boy-sta%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

                    GameSpot

                    Image
                    body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分
                    Read more

                    日野市

                    Image
                    body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 日野市 日野市 日文轉寫  • 日文 日野市  • 平假名 ひのし  • 罗马字 Hino-shi 金剛寺（關東三十六不動尊靈場第9號）的不動堂 市旗 徽章 日野市在東京都的位置 日野市 日野市在日本的位置 坐标： 35°40′17″N 139°23′42″E  /  35.6714°N 139.395°E  / 35.6714; 139.395 国家   日本 地方 關東地方 都道府縣 東京都 接鄰行政區 府中市、國立市、立川市、昭島市、八王子市、多摩市 政府  • 市長 大坪冬彦 面积  •  总计 27.53 平方公里（10.63 平方英里） 人口 （2014年12月1日）  • 總計 183,323  • 密度 6,660/平方公里（17,200/平方英里） 象徵  • 市樹 樫  • 市花 菊花  • 市鳥 普通翠鳥 时区 日本標準時間 （UTC+9） 地方公共團體編號 13212-8 邮政编码 〒 191-8686 市役所地址 日野市神明一丁目12番1號 電話號碼 +81-42-585-1111 法人編號 1000020132128 網站 http://www.city.hino.lg.jp/ 人口：日野市官方網頁 日野市 （日语： 日野市 ／ ひのし   Hino shi   * / ? ）為一位于東京都（不含島嶼部分）中央地帶的城市。從東京站乘坐中央線特別快速列車45分鐘即可到該市。面積27.53km²。1963年（昭和38年）11月3日，該市開始實行市制，為全日本第559個市。汽車企業日野自動車的總部位於這裡。 往東京都特別區部的通勤率為20.9%，往八王子市的通勤率為12.0%（平成22年國勢調査）。 目录 1 概要
                    Read more

                    Tu-95轟炸機

                    Image
                    body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機
                    Read more