Number of functions with certain property












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$begingroup$


Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?



My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.






combinatorics permutations







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$endgroup$

















    0












    $begingroup$


    Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?



    My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.






    combinatorics permutations







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?



      My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.






      combinatorics permutations







      share|cite|improve this question











      $endgroup$




      Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?



      My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.






      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 25 at 10:58









      N. F. Taussig

      44.6k103357




      44.6k103357










      asked Jan 25 at 3:55









      user601297user601297

      36919




      36919






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.



          Hence the number of functions in this case is $3^3$.



          Remember $f$ is one specific function.



          It is easier to count the number of $g$s that do not equal $f$ anywhere.



          In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
          of such functions is $2^3$.



          Hence the number of the other $g$s, which is what you are looking for, is
          $3^3-2^3$.



          Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            By size do you mean the number of values $g$ takes ???
            $endgroup$
            – user601297
            Jan 25 at 4:11










          • $begingroup$
            @user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
            $endgroup$
            – copper.hat
            Jan 25 at 4:12












          • $begingroup$
            But the example that @Bram28 gave has $g$ take all 3 values.
            $endgroup$
            – user601297
            Jan 25 at 4:20










          • $begingroup$
            In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
            $endgroup$
            – copper.hat
            Jan 25 at 4:23












          • $begingroup$
            Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
            $endgroup$
            – user601297
            Jan 25 at 4:34



















          0












          $begingroup$

          It is more important to understand why your answer does not work.



          Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$



          Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.



          Do you see how you can correct for this?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
            $endgroup$
            – user601297
            Jan 25 at 4:03










          • $begingroup$
            @user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
            $endgroup$
            – Bram28
            Jan 25 at 4:04










          • $begingroup$
            I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
            $endgroup$
            – user601297
            Jan 25 at 4:06



















          0












          $begingroup$

          Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.



            Hence the number of functions in this case is $3^3$.



            Remember $f$ is one specific function.



            It is easier to count the number of $g$s that do not equal $f$ anywhere.



            In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
            of such functions is $2^3$.



            Hence the number of the other $g$s, which is what you are looking for, is
            $3^3-2^3$.



            Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              By size do you mean the number of values $g$ takes ???
              $endgroup$
              – user601297
              Jan 25 at 4:11










            • $begingroup$
              @user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:12












            • $begingroup$
              But the example that @Bram28 gave has $g$ take all 3 values.
              $endgroup$
              – user601297
              Jan 25 at 4:20










            • $begingroup$
              In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:23












            • $begingroup$
              Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
              $endgroup$
              – user601297
              Jan 25 at 4:34
















            2












            $begingroup$

            In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.



            Hence the number of functions in this case is $3^3$.



            Remember $f$ is one specific function.



            It is easier to count the number of $g$s that do not equal $f$ anywhere.



            In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
            of such functions is $2^3$.



            Hence the number of the other $g$s, which is what you are looking for, is
            $3^3-2^3$.



            Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              By size do you mean the number of values $g$ takes ???
              $endgroup$
              – user601297
              Jan 25 at 4:11










            • $begingroup$
              @user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:12












            • $begingroup$
              But the example that @Bram28 gave has $g$ take all 3 values.
              $endgroup$
              – user601297
              Jan 25 at 4:20










            • $begingroup$
              In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:23












            • $begingroup$
              Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
              $endgroup$
              – user601297
              Jan 25 at 4:34














            2












            2








            2





            $begingroup$

            In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.



            Hence the number of functions in this case is $3^3$.



            Remember $f$ is one specific function.



            It is easier to count the number of $g$s that do not equal $f$ anywhere.



            In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
            of such functions is $2^3$.



            Hence the number of the other $g$s, which is what you are looking for, is
            $3^3-2^3$.



            Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.






            share|cite|improve this answer











            $endgroup$



            In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.



            Hence the number of functions in this case is $3^3$.



            Remember $f$ is one specific function.



            It is easier to count the number of $g$s that do not equal $f$ anywhere.



            In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
            of such functions is $2^3$.



            Hence the number of the other $g$s, which is what you are looking for, is
            $3^3-2^3$.



            Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 25 at 4:25

























            answered Jan 25 at 4:06









            copper.hatcopper.hat

            127k559160




            127k559160












            • $begingroup$
              By size do you mean the number of values $g$ takes ???
              $endgroup$
              – user601297
              Jan 25 at 4:11










            • $begingroup$
              @user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:12












            • $begingroup$
              But the example that @Bram28 gave has $g$ take all 3 values.
              $endgroup$
              – user601297
              Jan 25 at 4:20










            • $begingroup$
              In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:23












            • $begingroup$
              Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
              $endgroup$
              – user601297
              Jan 25 at 4:34


















            • $begingroup$
              By size do you mean the number of values $g$ takes ???
              $endgroup$
              – user601297
              Jan 25 at 4:11










            • $begingroup$
              @user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:12












            • $begingroup$
              But the example that @Bram28 gave has $g$ take all 3 values.
              $endgroup$
              – user601297
              Jan 25 at 4:20










            • $begingroup$
              In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
              $endgroup$
              – copper.hat
              Jan 25 at 4:23












            • $begingroup$
              Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
              $endgroup$
              – user601297
              Jan 25 at 4:34
















            $begingroup$
            By size do you mean the number of values $g$ takes ???
            $endgroup$
            – user601297
            Jan 25 at 4:11




            $begingroup$
            By size do you mean the number of values $g$ takes ???
            $endgroup$
            – user601297
            Jan 25 at 4:11












            $begingroup$
            @user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
            $endgroup$
            – copper.hat
            Jan 25 at 4:12






            $begingroup$
            @user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
            $endgroup$
            – copper.hat
            Jan 25 at 4:12














            $begingroup$
            But the example that @Bram28 gave has $g$ take all 3 values.
            $endgroup$
            – user601297
            Jan 25 at 4:20




            $begingroup$
            But the example that @Bram28 gave has $g$ take all 3 values.
            $endgroup$
            – user601297
            Jan 25 at 4:20












            $begingroup$
            In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
            $endgroup$
            – copper.hat
            Jan 25 at 4:23






            $begingroup$
            In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
            $endgroup$
            – copper.hat
            Jan 25 at 4:23














            $begingroup$
            Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
            $endgroup$
            – user601297
            Jan 25 at 4:34




            $begingroup$
            Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
            $endgroup$
            – user601297
            Jan 25 at 4:34











            0












            $begingroup$

            It is more important to understand why your answer does not work.



            Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$



            Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.



            Do you see how you can correct for this?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
              $endgroup$
              – user601297
              Jan 25 at 4:03










            • $begingroup$
              @user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
              $endgroup$
              – Bram28
              Jan 25 at 4:04










            • $begingroup$
              I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
              $endgroup$
              – user601297
              Jan 25 at 4:06
















            0












            $begingroup$

            It is more important to understand why your answer does not work.



            Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$



            Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.



            Do you see how you can correct for this?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
              $endgroup$
              – user601297
              Jan 25 at 4:03










            • $begingroup$
              @user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
              $endgroup$
              – Bram28
              Jan 25 at 4:04










            • $begingroup$
              I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
              $endgroup$
              – user601297
              Jan 25 at 4:06














            0












            0








            0





            $begingroup$

            It is more important to understand why your answer does not work.



            Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$



            Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.



            Do you see how you can correct for this?






            share|cite|improve this answer









            $endgroup$



            It is more important to understand why your answer does not work.



            Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$



            Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.



            Do you see how you can correct for this?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 25 at 4:00









            Bram28Bram28

            63.3k44793




            63.3k44793












            • $begingroup$
              I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
              $endgroup$
              – user601297
              Jan 25 at 4:03










            • $begingroup$
              @user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
              $endgroup$
              – Bram28
              Jan 25 at 4:04










            • $begingroup$
              I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
              $endgroup$
              – user601297
              Jan 25 at 4:06


















            • $begingroup$
              I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
              $endgroup$
              – user601297
              Jan 25 at 4:03










            • $begingroup$
              @user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
              $endgroup$
              – Bram28
              Jan 25 at 4:04










            • $begingroup$
              I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
              $endgroup$
              – user601297
              Jan 25 at 4:06
















            $begingroup$
            I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
            $endgroup$
            – user601297
            Jan 25 at 4:03




            $begingroup$
            I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
            $endgroup$
            – user601297
            Jan 25 at 4:03












            $begingroup$
            @user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
            $endgroup$
            – Bram28
            Jan 25 at 4:04




            $begingroup$
            @user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
            $endgroup$
            – Bram28
            Jan 25 at 4:04












            $begingroup$
            I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
            $endgroup$
            – user601297
            Jan 25 at 4:06




            $begingroup$
            I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
            $endgroup$
            – user601297
            Jan 25 at 4:06











            0












            $begingroup$

            Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.






                share|cite|improve this answer









                $endgroup$



                Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 4:10









                Jens SchwaigerJens Schwaiger

                1,631138




                1,631138






























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