Finding total unique string combination $s(r, n) = sum_{k=1}^{r} 4^k {{n+k-2}choose{k}}$ for $1 leq r leq 4$...

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I have a problem in counting the total of unique string combination. Provided a string of length 5 which is composed of 4 characters (ABCD), I would like to generate all the possible string combinations from one to four characters addition excluding both end addition.



Provided that the input string is ABCDA, I want to insert additional characters: w; w and x; w, x and y; w, x, y and z into any possible combination into the input string while avoiding insertion at both ends. So that the resulting combination would be: (AwBCDA, ABwCDA, ABCwDA,..., ABCDwA); (AwxBCDA, AwBxCDA, AwBCxDA,..., ABCDwxA); (AwxyBCDA, AwxByCDA, AwxBCyDA,..., ABCDwxyA); (AwxyzBCDA, AwxyBzCDA, AwxyBCzDA,..., ABCDwxyzA). And then for each substituents w, x, y, and z are substituted into A, B, C, or D.



I solved the combination above with the formula:
$$
s(r, n) = sum_{k=1}^{r} 4^{k} {{n+k-2}choose{k}}
$$

for $1leq rleq4$. The result from the above formula contains duplicates and I would like to count all the unique strings in $s(r, n)$. I tried to solve it using the following double summation:
$$
s_{U}(r, n) = sum_{k=1}^{r}1+sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
$$

which is rearranged to:
$$
s_{U}(r, n) = r+sum_{k=1}^{r}sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
$$

for $1 leq r leq 4$. Please kindly check whether is the formula and also the writing is correct or not.



Thank you.










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    1












    $begingroup$


    I have a problem in counting the total of unique string combination. Provided a string of length 5 which is composed of 4 characters (ABCD), I would like to generate all the possible string combinations from one to four characters addition excluding both end addition.



    Provided that the input string is ABCDA, I want to insert additional characters: w; w and x; w, x and y; w, x, y and z into any possible combination into the input string while avoiding insertion at both ends. So that the resulting combination would be: (AwBCDA, ABwCDA, ABCwDA,..., ABCDwA); (AwxBCDA, AwBxCDA, AwBCxDA,..., ABCDwxA); (AwxyBCDA, AwxByCDA, AwxBCyDA,..., ABCDwxyA); (AwxyzBCDA, AwxyBzCDA, AwxyBCzDA,..., ABCDwxyzA). And then for each substituents w, x, y, and z are substituted into A, B, C, or D.



    I solved the combination above with the formula:
    $$
    s(r, n) = sum_{k=1}^{r} 4^{k} {{n+k-2}choose{k}}
    $$

    for $1leq rleq4$. The result from the above formula contains duplicates and I would like to count all the unique strings in $s(r, n)$. I tried to solve it using the following double summation:
    $$
    s_{U}(r, n) = sum_{k=1}^{r}1+sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
    $$

    which is rearranged to:
    $$
    s_{U}(r, n) = r+sum_{k=1}^{r}sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
    $$

    for $1 leq r leq 4$. Please kindly check whether is the formula and also the writing is correct or not.



    Thank you.










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have a problem in counting the total of unique string combination. Provided a string of length 5 which is composed of 4 characters (ABCD), I would like to generate all the possible string combinations from one to four characters addition excluding both end addition.



      Provided that the input string is ABCDA, I want to insert additional characters: w; w and x; w, x and y; w, x, y and z into any possible combination into the input string while avoiding insertion at both ends. So that the resulting combination would be: (AwBCDA, ABwCDA, ABCwDA,..., ABCDwA); (AwxBCDA, AwBxCDA, AwBCxDA,..., ABCDwxA); (AwxyBCDA, AwxByCDA, AwxBCyDA,..., ABCDwxyA); (AwxyzBCDA, AwxyBzCDA, AwxyBCzDA,..., ABCDwxyzA). And then for each substituents w, x, y, and z are substituted into A, B, C, or D.



      I solved the combination above with the formula:
      $$
      s(r, n) = sum_{k=1}^{r} 4^{k} {{n+k-2}choose{k}}
      $$

      for $1leq rleq4$. The result from the above formula contains duplicates and I would like to count all the unique strings in $s(r, n)$. I tried to solve it using the following double summation:
      $$
      s_{U}(r, n) = sum_{k=1}^{r}1+sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
      $$

      which is rearranged to:
      $$
      s_{U}(r, n) = r+sum_{k=1}^{r}sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
      $$

      for $1 leq r leq 4$. Please kindly check whether is the formula and also the writing is correct or not.



      Thank you.










      share|cite|improve this question











      $endgroup$




      I have a problem in counting the total of unique string combination. Provided a string of length 5 which is composed of 4 characters (ABCD), I would like to generate all the possible string combinations from one to four characters addition excluding both end addition.



      Provided that the input string is ABCDA, I want to insert additional characters: w; w and x; w, x and y; w, x, y and z into any possible combination into the input string while avoiding insertion at both ends. So that the resulting combination would be: (AwBCDA, ABwCDA, ABCwDA,..., ABCDwA); (AwxBCDA, AwBxCDA, AwBCxDA,..., ABCDwxA); (AwxyBCDA, AwxByCDA, AwxBCyDA,..., ABCDwxyA); (AwxyzBCDA, AwxyBzCDA, AwxyBCzDA,..., ABCDwxyzA). And then for each substituents w, x, y, and z are substituted into A, B, C, or D.



      I solved the combination above with the formula:
      $$
      s(r, n) = sum_{k=1}^{r} 4^{k} {{n+k-2}choose{k}}
      $$

      for $1leq rleq4$. The result from the above formula contains duplicates and I would like to count all the unique strings in $s(r, n)$. I tried to solve it using the following double summation:
      $$
      s_{U}(r, n) = sum_{k=1}^{r}1+sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
      $$

      which is rearranged to:
      $$
      s_{U}(r, n) = r+sum_{k=1}^{r}sum_{l=1}^{k}3^{l}{{n+k-2}choose {l}}
      $$

      for $1 leq r leq 4$. Please kindly check whether is the formula and also the writing is correct or not.



      Thank you.







      combinatorics proof-verification






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      edited Jan 28 at 20:34









      Alex Ravsky

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      42.8k32483










      asked Jan 9 at 7:00









      Vic BrownVic Brown

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