How to prove $ frac{m!}{n!} geq n^{m-n} $












4












$begingroup$


How to prove the following:



$$ frac{m!}{n!} geq n^{m-n} $$



In my book it's written: "easy to prove by separately considering the cases $m geq n$ and $m<n$).



I tried using the bounds of Stirling and I got:



$$ frac{m!}{n!} geq frac{sqrt{2pi}}{e} n^{m-n} $$



But this bound is not tight as the first since $frac{sqrt{2pi}}{e}approx 0.92$



Thanks!






inequality factorial







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  • $begingroup$
    Closing this question? I don't understand...
    $endgroup$
    – Robert Z
    Jan 25 at 17:47
















4












$begingroup$


How to prove the following:



$$ frac{m!}{n!} geq n^{m-n} $$



In my book it's written: "easy to prove by separately considering the cases $m geq n$ and $m<n$).



I tried using the bounds of Stirling and I got:



$$ frac{m!}{n!} geq frac{sqrt{2pi}}{e} n^{m-n} $$



But this bound is not tight as the first since $frac{sqrt{2pi}}{e}approx 0.92$



Thanks!






inequality factorial







share|cite|improve this question











$endgroup$












  • $begingroup$
    Closing this question? I don't understand...
    $endgroup$
    – Robert Z
    Jan 25 at 17:47














4












4








4





$begingroup$


How to prove the following:



$$ frac{m!}{n!} geq n^{m-n} $$



In my book it's written: "easy to prove by separately considering the cases $m geq n$ and $m<n$).



I tried using the bounds of Stirling and I got:



$$ frac{m!}{n!} geq frac{sqrt{2pi}}{e} n^{m-n} $$



But this bound is not tight as the first since $frac{sqrt{2pi}}{e}approx 0.92$



Thanks!






inequality factorial







share|cite|improve this question











$endgroup$




How to prove the following:



$$ frac{m!}{n!} geq n^{m-n} $$



In my book it's written: "easy to prove by separately considering the cases $m geq n$ and $m<n$).



I tried using the bounds of Stirling and I got:



$$ frac{m!}{n!} geq frac{sqrt{2pi}}{e} n^{m-n} $$



But this bound is not tight as the first since $frac{sqrt{2pi}}{e}approx 0.92$



Thanks!






inequality factorial




inequality factorial






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share|cite|improve this question













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edited Jan 26 at 7:44









user21820

39.7k543157




39.7k543157










asked Jan 25 at 17:24









FelipeFelipe

12510




12510












  • $begingroup$
    Closing this question? I don't understand...
    $endgroup$
    – Robert Z
    Jan 25 at 17:47


















  • $begingroup$
    Closing this question? I don't understand...
    $endgroup$
    – Robert Z
    Jan 25 at 17:47
















$begingroup$
Closing this question? I don't understand...
$endgroup$
– Robert Z
Jan 25 at 17:47




$begingroup$
Closing this question? I don't understand...
$endgroup$
– Robert Z
Jan 25 at 17:47










4 Answers
4






active

oldest

votes


















10












$begingroup$

Stirling approximation is not useful here. The definition of factorial is all we need.



Note that if $mgeq n$ then
$$m!=underbrace{mcdot (m-1)cdots (n+1)}_{text{$m-n$ factors each one $>n$}}cdot n!$$
On the other hand if $n>m$ then
$$n!=underbrace{ncdot (n-1)cdots (m+1)}_{text{$n-m$ factors each one $leq n$}}cdot m!$$
Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I upvoted you Robert! Thanks!
    $endgroup$
    – Felipe
    Jan 25 at 17:55












  • $begingroup$
    Thanks. You are welcome!
    $endgroup$
    – Robert Z
    Jan 25 at 17:57





















4












$begingroup$

For $m ge n$
$$frac{m!}{n!}=(n+1) cdot (n+2) cdots (m-1) cdot m ge n cdot n cdots n = n^{m-n}$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Hint:



    For $m ge n$,
    $$frac{m!}{n!} = mtimes(m-1)timescdotstimes(n+1) ge ntimes n times cdots n=n^{m-n}$$
    What happens for $m le n$?
    $$frac{m!}{n!} = frac{1}{ntimes(n-1)timescdotstimes(m+1)} ge ?$$






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      No need for Stirling here, elementary computationw work way better.



      When $n leq m$, $m!/n!$ is a product of $n-m$ integers, all greater than $n$, thus $m! geq n^{m-n}n!$.



      When $n > m$, $m!/n!$ is a product of reciprocals of $n-m$ integers all $leq n$, so the product is $geq (1/n)^{n-m}=n^{m-n}$.






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        10












        $begingroup$

        Stirling approximation is not useful here. The definition of factorial is all we need.



        Note that if $mgeq n$ then
        $$m!=underbrace{mcdot (m-1)cdots (n+1)}_{text{$m-n$ factors each one $>n$}}cdot n!$$
        On the other hand if $n>m$ then
        $$n!=underbrace{ncdot (n-1)cdots (m+1)}_{text{$n-m$ factors each one $leq n$}}cdot m!$$
        Can you take it from here?






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          I upvoted you Robert! Thanks!
          $endgroup$
          – Felipe
          Jan 25 at 17:55












        • $begingroup$
          Thanks. You are welcome!
          $endgroup$
          – Robert Z
          Jan 25 at 17:57


















        10












        $begingroup$

        Stirling approximation is not useful here. The definition of factorial is all we need.



        Note that if $mgeq n$ then
        $$m!=underbrace{mcdot (m-1)cdots (n+1)}_{text{$m-n$ factors each one $>n$}}cdot n!$$
        On the other hand if $n>m$ then
        $$n!=underbrace{ncdot (n-1)cdots (m+1)}_{text{$n-m$ factors each one $leq n$}}cdot m!$$
        Can you take it from here?






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          I upvoted you Robert! Thanks!
          $endgroup$
          – Felipe
          Jan 25 at 17:55












        • $begingroup$
          Thanks. You are welcome!
          $endgroup$
          – Robert Z
          Jan 25 at 17:57
















        10












        10








        10





        $begingroup$

        Stirling approximation is not useful here. The definition of factorial is all we need.



        Note that if $mgeq n$ then
        $$m!=underbrace{mcdot (m-1)cdots (n+1)}_{text{$m-n$ factors each one $>n$}}cdot n!$$
        On the other hand if $n>m$ then
        $$n!=underbrace{ncdot (n-1)cdots (m+1)}_{text{$n-m$ factors each one $leq n$}}cdot m!$$
        Can you take it from here?






        share|cite|improve this answer











        $endgroup$



        Stirling approximation is not useful here. The definition of factorial is all we need.



        Note that if $mgeq n$ then
        $$m!=underbrace{mcdot (m-1)cdots (n+1)}_{text{$m-n$ factors each one $>n$}}cdot n!$$
        On the other hand if $n>m$ then
        $$n!=underbrace{ncdot (n-1)cdots (m+1)}_{text{$n-m$ factors each one $leq n$}}cdot m!$$
        Can you take it from here?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 25 at 17:34

























        answered Jan 25 at 17:27









        Robert ZRobert Z

        101k1070143




        101k1070143












        • $begingroup$
          I upvoted you Robert! Thanks!
          $endgroup$
          – Felipe
          Jan 25 at 17:55












        • $begingroup$
          Thanks. You are welcome!
          $endgroup$
          – Robert Z
          Jan 25 at 17:57




















        • $begingroup$
          I upvoted you Robert! Thanks!
          $endgroup$
          – Felipe
          Jan 25 at 17:55












        • $begingroup$
          Thanks. You are welcome!
          $endgroup$
          – Robert Z
          Jan 25 at 17:57


















        $begingroup$
        I upvoted you Robert! Thanks!
        $endgroup$
        – Felipe
        Jan 25 at 17:55






        $begingroup$
        I upvoted you Robert! Thanks!
        $endgroup$
        – Felipe
        Jan 25 at 17:55














        $begingroup$
        Thanks. You are welcome!
        $endgroup$
        – Robert Z
        Jan 25 at 17:57






        $begingroup$
        Thanks. You are welcome!
        $endgroup$
        – Robert Z
        Jan 25 at 17:57













        4












        $begingroup$

        For $m ge n$
        $$frac{m!}{n!}=(n+1) cdot (n+2) cdots (m-1) cdot m ge n cdot n cdots n = n^{m-n}$$






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          For $m ge n$
          $$frac{m!}{n!}=(n+1) cdot (n+2) cdots (m-1) cdot m ge n cdot n cdots n = n^{m-n}$$






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            For $m ge n$
            $$frac{m!}{n!}=(n+1) cdot (n+2) cdots (m-1) cdot m ge n cdot n cdots n = n^{m-n}$$






            share|cite|improve this answer









            $endgroup$



            For $m ge n$
            $$frac{m!}{n!}=(n+1) cdot (n+2) cdots (m-1) cdot m ge n cdot n cdots n = n^{m-n}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 25 at 17:27









            CrostulCrostul

            28.2k22352




            28.2k22352























                3












                $begingroup$

                Hint:



                For $m ge n$,
                $$frac{m!}{n!} = mtimes(m-1)timescdotstimes(n+1) ge ntimes n times cdots n=n^{m-n}$$
                What happens for $m le n$?
                $$frac{m!}{n!} = frac{1}{ntimes(n-1)timescdotstimes(m+1)} ge ?$$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Hint:



                  For $m ge n$,
                  $$frac{m!}{n!} = mtimes(m-1)timescdotstimes(n+1) ge ntimes n times cdots n=n^{m-n}$$
                  What happens for $m le n$?
                  $$frac{m!}{n!} = frac{1}{ntimes(n-1)timescdotstimes(m+1)} ge ?$$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Hint:



                    For $m ge n$,
                    $$frac{m!}{n!} = mtimes(m-1)timescdotstimes(n+1) ge ntimes n times cdots n=n^{m-n}$$
                    What happens for $m le n$?
                    $$frac{m!}{n!} = frac{1}{ntimes(n-1)timescdotstimes(m+1)} ge ?$$






                    share|cite|improve this answer









                    $endgroup$



                    Hint:



                    For $m ge n$,
                    $$frac{m!}{n!} = mtimes(m-1)timescdotstimes(n+1) ge ntimes n times cdots n=n^{m-n}$$
                    What happens for $m le n$?
                    $$frac{m!}{n!} = frac{1}{ntimes(n-1)timescdotstimes(m+1)} ge ?$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 25 at 17:27









                    Math LoverMath Lover

                    14.1k31437




                    14.1k31437























                        3












                        $begingroup$

                        No need for Stirling here, elementary computationw work way better.



                        When $n leq m$, $m!/n!$ is a product of $n-m$ integers, all greater than $n$, thus $m! geq n^{m-n}n!$.



                        When $n > m$, $m!/n!$ is a product of reciprocals of $n-m$ integers all $leq n$, so the product is $geq (1/n)^{n-m}=n^{m-n}$.






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          No need for Stirling here, elementary computationw work way better.



                          When $n leq m$, $m!/n!$ is a product of $n-m$ integers, all greater than $n$, thus $m! geq n^{m-n}n!$.



                          When $n > m$, $m!/n!$ is a product of reciprocals of $n-m$ integers all $leq n$, so the product is $geq (1/n)^{n-m}=n^{m-n}$.






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            No need for Stirling here, elementary computationw work way better.



                            When $n leq m$, $m!/n!$ is a product of $n-m$ integers, all greater than $n$, thus $m! geq n^{m-n}n!$.



                            When $n > m$, $m!/n!$ is a product of reciprocals of $n-m$ integers all $leq n$, so the product is $geq (1/n)^{n-m}=n^{m-n}$.






                            share|cite|improve this answer









                            $endgroup$



                            No need for Stirling here, elementary computationw work way better.



                            When $n leq m$, $m!/n!$ is a product of $n-m$ integers, all greater than $n$, thus $m! geq n^{m-n}n!$.



                            When $n > m$, $m!/n!$ is a product of reciprocals of $n-m$ integers all $leq n$, so the product is $geq (1/n)^{n-m}=n^{m-n}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 25 at 17:28









                            MindlackMindlack

                            4,920211




                            4,920211






























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