Solving equations using permutations and combinations [closed]












0












$begingroup$


I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$






combinatorics permutations







share|cite|improve this question











$endgroup$



closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
    $endgroup$
    – tommycautero
    Jan 14 at 18:17
















0












$begingroup$


I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$






combinatorics permutations







share|cite|improve this question











$endgroup$



closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
    $endgroup$
    – tommycautero
    Jan 14 at 18:17














0












0








0





$begingroup$


I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$






combinatorics permutations







share|cite|improve this question











$endgroup$




I need to solve for $n$ if $$P^n_4=60times C^n_2$$I’ve gotten up to $$n(n-1)(n-2)(n-3)=30n(n-1)$$






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 Jan 14 at 18:13









Shubham Johri

5,412818




5,412818










asked Jan 14 at 17:53









Sarah Sarah

31




31




closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by darij grinberg, Namaste, Holo, user91500, José Carlos Santos Jan 15 at 9:34


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
    $endgroup$
    – tommycautero
    Jan 14 at 18:17


















  • $begingroup$
    simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
    $endgroup$
    – tommycautero
    Jan 14 at 18:17
















$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17




$begingroup$
simplify n(n-1) on both sides, you get (n-2)(n-3)=30, which leads to solution n=8 (or n=-3, which i suppose makes no sense because you are working with n positive)
$endgroup$
– tommycautero
Jan 14 at 18:17










3 Answers
3






active

oldest

votes


















1












$begingroup$

Now your equation can be simplified to



$n(n-1)[(n-2)(n-3)-30]=0$
which can be further simplified to



$n(n-1)[n^2-5n-24]=0$



From here we get following values



$n=-3,0,1,8$



Now since $ngeq4$



$n=8$ is your answer.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.






      share|cite|improve this answer









      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Now your equation can be simplified to



        $n(n-1)[(n-2)(n-3)-30]=0$
        which can be further simplified to



        $n(n-1)[n^2-5n-24]=0$



        From here we get following values



        $n=-3,0,1,8$



        Now since $ngeq4$



        $n=8$ is your answer.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Now your equation can be simplified to



          $n(n-1)[(n-2)(n-3)-30]=0$
          which can be further simplified to



          $n(n-1)[n^2-5n-24]=0$



          From here we get following values



          $n=-3,0,1,8$



          Now since $ngeq4$



          $n=8$ is your answer.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Now your equation can be simplified to



            $n(n-1)[(n-2)(n-3)-30]=0$
            which can be further simplified to



            $n(n-1)[n^2-5n-24]=0$



            From here we get following values



            $n=-3,0,1,8$



            Now since $ngeq4$



            $n=8$ is your answer.






            share|cite|improve this answer









            $endgroup$



            Now your equation can be simplified to



            $n(n-1)[(n-2)(n-3)-30]=0$
            which can be further simplified to



            $n(n-1)[n^2-5n-24]=0$



            From here we get following values



            $n=-3,0,1,8$



            Now since $ngeq4$



            $n=8$ is your answer.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 14 at 18:16









            JimmyJimmy

            30813




            30813























                1












                $begingroup$

                By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.






                    share|cite|improve this answer









                    $endgroup$



                    By eliminating common terms on both sides, you get $(n-2)(n-3)=30$, or $n^2-5n-24=0$. Positive solution is $n=8$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 14 at 18:18









                    herb steinbergherb steinberg

                    3,0982311




                    3,0982311























                        0












                        $begingroup$

                        You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.






                            share|cite|improve this answer









                            $endgroup$



                            You are nearly there! You got$$n(n-1)(n-2)(n-3)=30n(n-1)$$Now shift all the terms to one side to get$$n(n-1)Big[(n-2)(n-3)-30Big]=0$$So $n$ could be $0,1$ or such that$$(n-2)(n-3)-30=0\implies n^2-5n-24=(n-8)(n+3)=0\therefore n=0,1,-3,8$$If you define $P^n_4=displaystylefrac{n!}{(n-4)!}$, you need $nge4$. This leaves you with the answer $n=8$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 14 at 18:19









                            Shubham JohriShubham Johri

                            5,412818




                            5,412818















                                -

                                Popular posts from this blog

                                Questions related to Moebius Transform of Characteristic Function of the Primes

                                Image
                                2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
                                Read more

                                List of scandals in India

                                Image
                                List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
                                Read more

                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                Image
                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                                Read more