Combinations - Solving recurrence relation in 2 variables without using generating functions

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I know that there are similar questions here solving 2 variable recurrence and combinations recurrence equation, but both of them use generating functions to solve this. Is there any other way to solve this problem:



$$Psi(n,k) = begincases
0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n = 0 text and
k > 0\
1 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n geq 0 text and
k=0\
Psi(n-1,k) + Psi(n-1,k-1);;text if n > 0 text and k > 0\
0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text otherwise
endcases$$



I know the solution is $binomnk$, but I would like to know a way to solve this without generating functions.




combinatorics discrete-mathematics recurrence-relations combinations generating-functions




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    up vote
    1
    down vote

    favorite












    I know that there are similar questions here solving 2 variable recurrence and combinations recurrence equation, but both of them use generating functions to solve this. Is there any other way to solve this problem:



    $$Psi(n,k) = begincases
    0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n = 0 text and
    k > 0\
    1 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n geq 0 text and
    k=0\
    Psi(n-1,k) + Psi(n-1,k-1);;text if n > 0 text and k > 0\
    0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text otherwise
    endcases$$



    I know the solution is $binomnk$, but I would like to know a way to solve this without generating functions.




    combinatorics discrete-mathematics recurrence-relations combinations generating-functions




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I know that there are similar questions here solving 2 variable recurrence and combinations recurrence equation, but both of them use generating functions to solve this. Is there any other way to solve this problem:



      $$Psi(n,k) = begincases
      0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n = 0 text and
      k > 0\
      1 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n geq 0 text and
      k=0\
      Psi(n-1,k) + Psi(n-1,k-1);;text if n > 0 text and k > 0\
      0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text otherwise
      endcases$$



      I know the solution is $binomnk$, but I would like to know a way to solve this without generating functions.




      combinatorics discrete-mathematics recurrence-relations combinations generating-functions




      share|cite|improve this question











      I know that there are similar questions here solving 2 variable recurrence and combinations recurrence equation, but both of them use generating functions to solve this. Is there any other way to solve this problem:



      $$Psi(n,k) = begincases
      0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n = 0 text and
      k > 0\
      1 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text if n geq 0 text and
      k=0\
      Psi(n-1,k) + Psi(n-1,k-1);;text if n > 0 text and k > 0\
      0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;text otherwise
      endcases$$



      I know the solution is $binomnk$, but I would like to know a way to solve this without generating functions.





      combinatorics discrete-mathematics recurrence-relations combinations generating-functions





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      asked Jul 20 at 9:31









      pedroth

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          The most natural way to solve this is the following: I'd start by making a table up to $n=5$ or so and then say "Ah, these are just the binomial coefficients". Then I'd set up an induction proof that this is indeed the case.






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          • I like your approach, but I would like one where the mathematician didn't know what are binomial coefficients
            – pedroth
            Jul 20 at 9:59










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          up vote
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          down vote













          The most natural way to solve this is the following: I'd start by making a table up to $n=5$ or so and then say "Ah, these are just the binomial coefficients". Then I'd set up an induction proof that this is indeed the case.






          share|cite|improve this answer





















          • I like your approach, but I would like one where the mathematician didn't know what are binomial coefficients
            – pedroth
            Jul 20 at 9:59














          up vote
          1
          down vote













          The most natural way to solve this is the following: I'd start by making a table up to $n=5$ or so and then say "Ah, these are just the binomial coefficients". Then I'd set up an induction proof that this is indeed the case.






          share|cite|improve this answer





















          • I like your approach, but I would like one where the mathematician didn't know what are binomial coefficients
            – pedroth
            Jul 20 at 9:59












          up vote
          1
          down vote










          up vote
          1
          down vote









          The most natural way to solve this is the following: I'd start by making a table up to $n=5$ or so and then say "Ah, these are just the binomial coefficients". Then I'd set up an induction proof that this is indeed the case.






          share|cite|improve this answer













          The most natural way to solve this is the following: I'd start by making a table up to $n=5$ or so and then say "Ah, these are just the binomial coefficients". Then I'd set up an induction proof that this is indeed the case.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 20 at 9:38









          Christian Blatter

          163k7107306




          163k7107306











          • I like your approach, but I would like one where the mathematician didn't know what are binomial coefficients
            – pedroth
            Jul 20 at 9:59
















          • I like your approach, but I would like one where the mathematician didn't know what are binomial coefficients
            – pedroth
            Jul 20 at 9:59















          I like your approach, but I would like one where the mathematician didn't know what are binomial coefficients
          – pedroth
          Jul 20 at 9:59




          I like your approach, but I would like one where the mathematician didn't know what are binomial coefficients
          – pedroth
          Jul 20 at 9:59












           

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