simple combinatorics: product rule and number of ordered pairs












0












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Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.



${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.



What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?






combinatorics







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    0












    $begingroup$


    Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.



    ${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.



    What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?






    combinatorics







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.



      ${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.



      What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?






      combinatorics







      share|cite|improve this question









      $endgroup$




      Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.



      ${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.



      What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?






      combinatorics




      combinatorics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 29 '18 at 0:02









      Just do itJust do it

      19618




      19618






















          1 Answer
          1






          active

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          1












          $begingroup$

          In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.



          In the second scenario, there are three cases:




          • Two elements from $X$

          • Two elements from $Y$

          • One element from $X$, one element from $Y$


          This is why there are more pairs in the second scenario than the first.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ou, oh, ye! Thx :D
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:06






          • 1




            $begingroup$
            It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
            $endgroup$
            – Henry
            Dec 29 '18 at 0:19












          • $begingroup$
            @Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:38










          • $begingroup$
            @Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:46













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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.



          In the second scenario, there are three cases:




          • Two elements from $X$

          • Two elements from $Y$

          • One element from $X$, one element from $Y$


          This is why there are more pairs in the second scenario than the first.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ou, oh, ye! Thx :D
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:06






          • 1




            $begingroup$
            It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
            $endgroup$
            – Henry
            Dec 29 '18 at 0:19












          • $begingroup$
            @Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:38










          • $begingroup$
            @Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:46


















          1












          $begingroup$

          In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.



          In the second scenario, there are three cases:




          • Two elements from $X$

          • Two elements from $Y$

          • One element from $X$, one element from $Y$


          This is why there are more pairs in the second scenario than the first.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ou, oh, ye! Thx :D
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:06






          • 1




            $begingroup$
            It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
            $endgroup$
            – Henry
            Dec 29 '18 at 0:19












          • $begingroup$
            @Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:38










          • $begingroup$
            @Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:46
















          1












          1








          1





          $begingroup$

          In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.



          In the second scenario, there are three cases:




          • Two elements from $X$

          • Two elements from $Y$

          • One element from $X$, one element from $Y$


          This is why there are more pairs in the second scenario than the first.






          share|cite|improve this answer









          $endgroup$



          In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.



          In the second scenario, there are three cases:




          • Two elements from $X$

          • Two elements from $Y$

          • One element from $X$, one element from $Y$


          This is why there are more pairs in the second scenario than the first.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 29 '18 at 0:05









          Noble MushtakNoble Mushtak

          15.3k1835




          15.3k1835












          • $begingroup$
            Ou, oh, ye! Thx :D
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:06






          • 1




            $begingroup$
            It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
            $endgroup$
            – Henry
            Dec 29 '18 at 0:19












          • $begingroup$
            @Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:38










          • $begingroup$
            @Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:46




















          • $begingroup$
            Ou, oh, ye! Thx :D
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:06






          • 1




            $begingroup$
            It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
            $endgroup$
            – Henry
            Dec 29 '18 at 0:19












          • $begingroup$
            @Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:38










          • $begingroup$
            @Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
            $endgroup$
            – Just do it
            Dec 29 '18 at 0:46


















          $begingroup$
          Ou, oh, ye! Thx :D
          $endgroup$
          – Just do it
          Dec 29 '18 at 0:06




          $begingroup$
          Ou, oh, ye! Thx :D
          $endgroup$
          – Just do it
          Dec 29 '18 at 0:06




          1




          1




          $begingroup$
          It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
          $endgroup$
          – Henry
          Dec 29 '18 at 0:19






          $begingroup$
          It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
          $endgroup$
          – Henry
          Dec 29 '18 at 0:19














          $begingroup$
          @Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
          $endgroup$
          – Just do it
          Dec 29 '18 at 0:38




          $begingroup$
          @Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
          $endgroup$
          – Just do it
          Dec 29 '18 at 0:38












          $begingroup$
          @Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
          $endgroup$
          – Just do it
          Dec 29 '18 at 0:46






          $begingroup$
          @Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
          $endgroup$
          – Just do it
          Dec 29 '18 at 0:46




















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