Counting unique element output from a product of combination












1












$begingroup$


I have a string of length n which is comprised of four characters only (say ABCD) and I would like to make 1 until 4 substitutions to this string. The character use for substitution would also comprised the same character ABCD. Therefore, the total ways for making such combination can be calculated:
$$
s^r binom{n}{r}
$$



where: $n$ = length of a string; $s$ = number of substituting characters (4); $r$ = number of position to be substituted in a string (1 to 4 position)



so for $n$ = 5 (say ABCDA), it will give:
$$
1subs = 4^1 binom{5}{1} = 20
$$
$$
2subs = 4^2 binom{5}{2} = 160
$$
$$
3subs = 4^3 binom{5}{3} = 640
$$
$$
4subs = 4^4 binom{5}{4} = 1280
$$



Example from 1subs:
ABCDA, BBCDA, CBCDA, DBCDA
AACDA, ABCDA, ACCDA, ADCDA
ABADA, ABBDA, ABCDA, ABDDA
ABCAA, ABCBA, ABCCA, ABCDA
ABCDA, ABCDB, ABCDC, ABCDD



Notice that ABCDA was duplicated 5 times and among the 20 elements, only 16 are unique. I currently solve the unique sets of these combinations using polynomial functions and I got:
$$
1subs = 3n + 1
$$
$$
2subs = frac{9}{2}n^2 - frac{3}{2}n + 1
$$
$$
3subs = frac{27}{6}n^3 - 9n^2 + 15n + 1
$$
$$
4subs = frac{27}{8}n^4 - frac{63}{4}n^3 + frac{225}{8}n^2 - frac{51}{4}n + 1
$$



So, for $n = 5$, the unique elements for 1 until 4 substitution will be: $16, 106, 376, 781$. However, the polynomial functions above are generated in the presence of data and I want to count the unique combinations for any arbitrary $n$ (assuming the absence of such data). Is there any combinatorial formula to solve this?



Thanks






combinatorics







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have a string of length n which is comprised of four characters only (say ABCD) and I would like to make 1 until 4 substitutions to this string. The character use for substitution would also comprised the same character ABCD. Therefore, the total ways for making such combination can be calculated:
    $$
    s^r binom{n}{r}
    $$



    where: $n$ = length of a string; $s$ = number of substituting characters (4); $r$ = number of position to be substituted in a string (1 to 4 position)



    so for $n$ = 5 (say ABCDA), it will give:
    $$
    1subs = 4^1 binom{5}{1} = 20
    $$
    $$
    2subs = 4^2 binom{5}{2} = 160
    $$
    $$
    3subs = 4^3 binom{5}{3} = 640
    $$
    $$
    4subs = 4^4 binom{5}{4} = 1280
    $$



    Example from 1subs:
    ABCDA, BBCDA, CBCDA, DBCDA
    AACDA, ABCDA, ACCDA, ADCDA
    ABADA, ABBDA, ABCDA, ABDDA
    ABCAA, ABCBA, ABCCA, ABCDA
    ABCDA, ABCDB, ABCDC, ABCDD



    Notice that ABCDA was duplicated 5 times and among the 20 elements, only 16 are unique. I currently solve the unique sets of these combinations using polynomial functions and I got:
    $$
    1subs = 3n + 1
    $$
    $$
    2subs = frac{9}{2}n^2 - frac{3}{2}n + 1
    $$
    $$
    3subs = frac{27}{6}n^3 - 9n^2 + 15n + 1
    $$
    $$
    4subs = frac{27}{8}n^4 - frac{63}{4}n^3 + frac{225}{8}n^2 - frac{51}{4}n + 1
    $$



    So, for $n = 5$, the unique elements for 1 until 4 substitution will be: $16, 106, 376, 781$. However, the polynomial functions above are generated in the presence of data and I want to count the unique combinations for any arbitrary $n$ (assuming the absence of such data). Is there any combinatorial formula to solve this?



    Thanks






    combinatorics







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have a string of length n which is comprised of four characters only (say ABCD) and I would like to make 1 until 4 substitutions to this string. The character use for substitution would also comprised the same character ABCD. Therefore, the total ways for making such combination can be calculated:
      $$
      s^r binom{n}{r}
      $$



      where: $n$ = length of a string; $s$ = number of substituting characters (4); $r$ = number of position to be substituted in a string (1 to 4 position)



      so for $n$ = 5 (say ABCDA), it will give:
      $$
      1subs = 4^1 binom{5}{1} = 20
      $$
      $$
      2subs = 4^2 binom{5}{2} = 160
      $$
      $$
      3subs = 4^3 binom{5}{3} = 640
      $$
      $$
      4subs = 4^4 binom{5}{4} = 1280
      $$



      Example from 1subs:
      ABCDA, BBCDA, CBCDA, DBCDA
      AACDA, ABCDA, ACCDA, ADCDA
      ABADA, ABBDA, ABCDA, ABDDA
      ABCAA, ABCBA, ABCCA, ABCDA
      ABCDA, ABCDB, ABCDC, ABCDD



      Notice that ABCDA was duplicated 5 times and among the 20 elements, only 16 are unique. I currently solve the unique sets of these combinations using polynomial functions and I got:
      $$
      1subs = 3n + 1
      $$
      $$
      2subs = frac{9}{2}n^2 - frac{3}{2}n + 1
      $$
      $$
      3subs = frac{27}{6}n^3 - 9n^2 + 15n + 1
      $$
      $$
      4subs = frac{27}{8}n^4 - frac{63}{4}n^3 + frac{225}{8}n^2 - frac{51}{4}n + 1
      $$



      So, for $n = 5$, the unique elements for 1 until 4 substitution will be: $16, 106, 376, 781$. However, the polynomial functions above are generated in the presence of data and I want to count the unique combinations for any arbitrary $n$ (assuming the absence of such data). Is there any combinatorial formula to solve this?



      Thanks






      combinatorics







      share|cite|improve this question











      $endgroup$




      I have a string of length n which is comprised of four characters only (say ABCD) and I would like to make 1 until 4 substitutions to this string. The character use for substitution would also comprised the same character ABCD. Therefore, the total ways for making such combination can be calculated:
      $$
      s^r binom{n}{r}
      $$



      where: $n$ = length of a string; $s$ = number of substituting characters (4); $r$ = number of position to be substituted in a string (1 to 4 position)



      so for $n$ = 5 (say ABCDA), it will give:
      $$
      1subs = 4^1 binom{5}{1} = 20
      $$
      $$
      2subs = 4^2 binom{5}{2} = 160
      $$
      $$
      3subs = 4^3 binom{5}{3} = 640
      $$
      $$
      4subs = 4^4 binom{5}{4} = 1280
      $$



      Example from 1subs:
      ABCDA, BBCDA, CBCDA, DBCDA
      AACDA, ABCDA, ACCDA, ADCDA
      ABADA, ABBDA, ABCDA, ABDDA
      ABCAA, ABCBA, ABCCA, ABCDA
      ABCDA, ABCDB, ABCDC, ABCDD



      Notice that ABCDA was duplicated 5 times and among the 20 elements, only 16 are unique. I currently solve the unique sets of these combinations using polynomial functions and I got:
      $$
      1subs = 3n + 1
      $$
      $$
      2subs = frac{9}{2}n^2 - frac{3}{2}n + 1
      $$
      $$
      3subs = frac{27}{6}n^3 - 9n^2 + 15n + 1
      $$
      $$
      4subs = frac{27}{8}n^4 - frac{63}{4}n^3 + frac{225}{8}n^2 - frac{51}{4}n + 1
      $$



      So, for $n = 5$, the unique elements for 1 until 4 substitution will be: $16, 106, 376, 781$. However, the polynomial functions above are generated in the presence of data and I want to count the unique combinations for any arbitrary $n$ (assuming the absence of such data). Is there any combinatorial formula to solve this?



      Thanks






      combinatorics




      combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 4:01







      Vic Brown

















      asked Dec 27 '18 at 3:48









      Vic BrownVic Brown

      213




      213






















          1 Answer
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          0












          $begingroup$

          The total number of strings of length $n$ comprised of four characters is $4^n$, that is $4$ options for each position in the string. To count the number of unique strings that can be result from the substitution of characters, count only the substitutions that change the string:
          $$3^rbinom{n}{r}$$
          This will give you the number of new strings that can be made when exactly $r$ characters have been replaced with a character other than the original. If you want to allow a character to be replaced with itself you need to take the sum for each $r$ up to the maximum number of characters to be replaced and include the original string:
          $$1+sum_{k=1}^r 3^kbinom{n}{k}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Wow..thanks a lot!! It really helps :D And, could you help to simplify the above formula by avoiding the sigma notation? So that I could calculate for cases involving large number of r and n without having to write each summation element recurrently.
            $endgroup$
            – Vic Brown
            Dec 27 '18 at 7:38











          Your Answer





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          0












          $begingroup$

          The total number of strings of length $n$ comprised of four characters is $4^n$, that is $4$ options for each position in the string. To count the number of unique strings that can be result from the substitution of characters, count only the substitutions that change the string:
          $$3^rbinom{n}{r}$$
          This will give you the number of new strings that can be made when exactly $r$ characters have been replaced with a character other than the original. If you want to allow a character to be replaced with itself you need to take the sum for each $r$ up to the maximum number of characters to be replaced and include the original string:
          $$1+sum_{k=1}^r 3^kbinom{n}{k}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Wow..thanks a lot!! It really helps :D And, could you help to simplify the above formula by avoiding the sigma notation? So that I could calculate for cases involving large number of r and n without having to write each summation element recurrently.
            $endgroup$
            – Vic Brown
            Dec 27 '18 at 7:38
















          0












          $begingroup$

          The total number of strings of length $n$ comprised of four characters is $4^n$, that is $4$ options for each position in the string. To count the number of unique strings that can be result from the substitution of characters, count only the substitutions that change the string:
          $$3^rbinom{n}{r}$$
          This will give you the number of new strings that can be made when exactly $r$ characters have been replaced with a character other than the original. If you want to allow a character to be replaced with itself you need to take the sum for each $r$ up to the maximum number of characters to be replaced and include the original string:
          $$1+sum_{k=1}^r 3^kbinom{n}{k}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Wow..thanks a lot!! It really helps :D And, could you help to simplify the above formula by avoiding the sigma notation? So that I could calculate for cases involving large number of r and n without having to write each summation element recurrently.
            $endgroup$
            – Vic Brown
            Dec 27 '18 at 7:38














          0












          0








          0





          $begingroup$

          The total number of strings of length $n$ comprised of four characters is $4^n$, that is $4$ options for each position in the string. To count the number of unique strings that can be result from the substitution of characters, count only the substitutions that change the string:
          $$3^rbinom{n}{r}$$
          This will give you the number of new strings that can be made when exactly $r$ characters have been replaced with a character other than the original. If you want to allow a character to be replaced with itself you need to take the sum for each $r$ up to the maximum number of characters to be replaced and include the original string:
          $$1+sum_{k=1}^r 3^kbinom{n}{k}$$






          share|cite|improve this answer









          $endgroup$



          The total number of strings of length $n$ comprised of four characters is $4^n$, that is $4$ options for each position in the string. To count the number of unique strings that can be result from the substitution of characters, count only the substitutions that change the string:
          $$3^rbinom{n}{r}$$
          This will give you the number of new strings that can be made when exactly $r$ characters have been replaced with a character other than the original. If you want to allow a character to be replaced with itself you need to take the sum for each $r$ up to the maximum number of characters to be replaced and include the original string:
          $$1+sum_{k=1}^r 3^kbinom{n}{k}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 4:32









          Daniel MathiasDaniel Mathias

          1,30518




          1,30518












          • $begingroup$
            Wow..thanks a lot!! It really helps :D And, could you help to simplify the above formula by avoiding the sigma notation? So that I could calculate for cases involving large number of r and n without having to write each summation element recurrently.
            $endgroup$
            – Vic Brown
            Dec 27 '18 at 7:38


















          • $begingroup$
            Wow..thanks a lot!! It really helps :D And, could you help to simplify the above formula by avoiding the sigma notation? So that I could calculate for cases involving large number of r and n without having to write each summation element recurrently.
            $endgroup$
            – Vic Brown
            Dec 27 '18 at 7:38
















          $begingroup$
          Wow..thanks a lot!! It really helps :D And, could you help to simplify the above formula by avoiding the sigma notation? So that I could calculate for cases involving large number of r and n without having to write each summation element recurrently.
          $endgroup$
          – Vic Brown
          Dec 27 '18 at 7:38




          $begingroup$
          Wow..thanks a lot!! It really helps :D And, could you help to simplify the above formula by avoiding the sigma notation? So that I could calculate for cases involving large number of r and n without having to write each summation element recurrently.
          $endgroup$
          – Vic Brown
          Dec 27 '18 at 7:38


















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