Why do I get two different answers for this counting problem?












2












$begingroup$


Suppose that we consider English alphabets so we have 26 letters which 5 of them are vowels. I want to find all three letters sequences that contain at least one vowel.



My first approach is $26^3-21^3=8315$ which is the number of all three lettres sequences minus the number of three letter sequences which do not contain vowels.



Second approach: at least one vowel means one vowel or two vowels or three vowels so the answer is $(5cdot21^2)+(5^2cdot21)+5^3=2855$.



Why are these two answers different?






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    2












    $begingroup$


    Suppose that we consider English alphabets so we have 26 letters which 5 of them are vowels. I want to find all three letters sequences that contain at least one vowel.



    My first approach is $26^3-21^3=8315$ which is the number of all three lettres sequences minus the number of three letter sequences which do not contain vowels.



    Second approach: at least one vowel means one vowel or two vowels or three vowels so the answer is $(5cdot21^2)+(5^2cdot21)+5^3=2855$.



    Why are these two answers different?






    combinatorics discrete-mathematics







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    New contributor




    Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      2












      2








      2





      $begingroup$


      Suppose that we consider English alphabets so we have 26 letters which 5 of them are vowels. I want to find all three letters sequences that contain at least one vowel.



      My first approach is $26^3-21^3=8315$ which is the number of all three lettres sequences minus the number of three letter sequences which do not contain vowels.



      Second approach: at least one vowel means one vowel or two vowels or three vowels so the answer is $(5cdot21^2)+(5^2cdot21)+5^3=2855$.



      Why are these two answers different?






      combinatorics discrete-mathematics







      share|cite|improve this question







      New contributor




      Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Suppose that we consider English alphabets so we have 26 letters which 5 of them are vowels. I want to find all three letters sequences that contain at least one vowel.



      My first approach is $26^3-21^3=8315$ which is the number of all three lettres sequences minus the number of three letter sequences which do not contain vowels.



      Second approach: at least one vowel means one vowel or two vowels or three vowels so the answer is $(5cdot21^2)+(5^2cdot21)+5^3=2855$.



      Why are these two answers different?






      combinatorics discrete-mathematics




      combinatorics discrete-mathematics






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      Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 2 hours ago









      StudentStudent

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          3 Answers
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          $begingroup$

          Take the number of all three letter words ($26^3$) and subtract from it the number that have only consonants ($21^3$).



          Your second method does not address all possible sequences of vowels, such as vowel-consonant-vowel or vowel-vowel-consonant or consonant-vowel-vowel.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That is my solution 1. I need to know why solution 2 does not work.
            $endgroup$
            – Student
            2 hours ago










          • $begingroup$
            I get it. Thanks so much
            $endgroup$
            – Student
            2 hours ago



















          2












          $begingroup$

          It is worth noting that to count properly with the second approach, you would have to write $$3 cdot 5 cdot 21^2 + 3 cdot 5^2 cdot 21 + 5^3 = 8315.$$ The first term counts all cases where there is exactly one vowel, and there are three positions for the location of this vowel. The second term counts all cases where there are exactly two vowels, and there are three positions for the location of the sole consonant. The third term counts all cases where there are exactly three vowels, but we do not multiply by three because all three letters are of the same type. You only have to consider the ordering of the types of letters (vowel vs. consonant) when there is more than one type of letter used.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Refer to Binomial Expansion



            $$(21+5)^3=21^3+tbinom 31cdotp 21^2cdotp5+tbinom 32cdotp21cdotp 5^2+5^3$$



            so$$26^3-21^3= (3cdot21^2cdotp 5)+(3cdot21cdotp 5^2)+5^3$$



            The binomial coefficients count the ways to select elements from a set.



            In your case, that is positions to place the vowels in the word.



            There are $3cdot21^2cdotp 5$ ways to select 1 vowels and two consonants where the vowel is in the first, second, or third place. And so on.






            share|cite|improve this answer









            $endgroup$














              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

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              active

              oldest

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              3












              $begingroup$

              Take the number of all three letter words ($26^3$) and subtract from it the number that have only consonants ($21^3$).



              Your second method does not address all possible sequences of vowels, such as vowel-consonant-vowel or vowel-vowel-consonant or consonant-vowel-vowel.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                That is my solution 1. I need to know why solution 2 does not work.
                $endgroup$
                – Student
                2 hours ago










              • $begingroup$
                I get it. Thanks so much
                $endgroup$
                – Student
                2 hours ago
















              3












              $begingroup$

              Take the number of all three letter words ($26^3$) and subtract from it the number that have only consonants ($21^3$).



              Your second method does not address all possible sequences of vowels, such as vowel-consonant-vowel or vowel-vowel-consonant or consonant-vowel-vowel.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                That is my solution 1. I need to know why solution 2 does not work.
                $endgroup$
                – Student
                2 hours ago










              • $begingroup$
                I get it. Thanks so much
                $endgroup$
                – Student
                2 hours ago














              3












              3








              3





              $begingroup$

              Take the number of all three letter words ($26^3$) and subtract from it the number that have only consonants ($21^3$).



              Your second method does not address all possible sequences of vowels, such as vowel-consonant-vowel or vowel-vowel-consonant or consonant-vowel-vowel.






              share|cite|improve this answer











              $endgroup$



              Take the number of all three letter words ($26^3$) and subtract from it the number that have only consonants ($21^3$).



              Your second method does not address all possible sequences of vowels, such as vowel-consonant-vowel or vowel-vowel-consonant or consonant-vowel-vowel.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 hours ago

























              answered 2 hours ago









              David G. StorkDavid G. Stork

              11.6k41534




              11.6k41534












              • $begingroup$
                That is my solution 1. I need to know why solution 2 does not work.
                $endgroup$
                – Student
                2 hours ago










              • $begingroup$
                I get it. Thanks so much
                $endgroup$
                – Student
                2 hours ago


















              • $begingroup$
                That is my solution 1. I need to know why solution 2 does not work.
                $endgroup$
                – Student
                2 hours ago










              • $begingroup$
                I get it. Thanks so much
                $endgroup$
                – Student
                2 hours ago
















              $begingroup$
              That is my solution 1. I need to know why solution 2 does not work.
              $endgroup$
              – Student
              2 hours ago




              $begingroup$
              That is my solution 1. I need to know why solution 2 does not work.
              $endgroup$
              – Student
              2 hours ago












              $begingroup$
              I get it. Thanks so much
              $endgroup$
              – Student
              2 hours ago




              $begingroup$
              I get it. Thanks so much
              $endgroup$
              – Student
              2 hours ago











              2












              $begingroup$

              It is worth noting that to count properly with the second approach, you would have to write $$3 cdot 5 cdot 21^2 + 3 cdot 5^2 cdot 21 + 5^3 = 8315.$$ The first term counts all cases where there is exactly one vowel, and there are three positions for the location of this vowel. The second term counts all cases where there are exactly two vowels, and there are three positions for the location of the sole consonant. The third term counts all cases where there are exactly three vowels, but we do not multiply by three because all three letters are of the same type. You only have to consider the ordering of the types of letters (vowel vs. consonant) when there is more than one type of letter used.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                It is worth noting that to count properly with the second approach, you would have to write $$3 cdot 5 cdot 21^2 + 3 cdot 5^2 cdot 21 + 5^3 = 8315.$$ The first term counts all cases where there is exactly one vowel, and there are three positions for the location of this vowel. The second term counts all cases where there are exactly two vowels, and there are three positions for the location of the sole consonant. The third term counts all cases where there are exactly three vowels, but we do not multiply by three because all three letters are of the same type. You only have to consider the ordering of the types of letters (vowel vs. consonant) when there is more than one type of letter used.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  It is worth noting that to count properly with the second approach, you would have to write $$3 cdot 5 cdot 21^2 + 3 cdot 5^2 cdot 21 + 5^3 = 8315.$$ The first term counts all cases where there is exactly one vowel, and there are three positions for the location of this vowel. The second term counts all cases where there are exactly two vowels, and there are three positions for the location of the sole consonant. The third term counts all cases where there are exactly three vowels, but we do not multiply by three because all three letters are of the same type. You only have to consider the ordering of the types of letters (vowel vs. consonant) when there is more than one type of letter used.






                  share|cite|improve this answer









                  $endgroup$



                  It is worth noting that to count properly with the second approach, you would have to write $$3 cdot 5 cdot 21^2 + 3 cdot 5^2 cdot 21 + 5^3 = 8315.$$ The first term counts all cases where there is exactly one vowel, and there are three positions for the location of this vowel. The second term counts all cases where there are exactly two vowels, and there are three positions for the location of the sole consonant. The third term counts all cases where there are exactly three vowels, but we do not multiply by three because all three letters are of the same type. You only have to consider the ordering of the types of letters (vowel vs. consonant) when there is more than one type of letter used.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  heropupheropup

                  64.9k764103




                  64.9k764103























                      1












                      $begingroup$

                      Refer to Binomial Expansion



                      $$(21+5)^3=21^3+tbinom 31cdotp 21^2cdotp5+tbinom 32cdotp21cdotp 5^2+5^3$$



                      so$$26^3-21^3= (3cdot21^2cdotp 5)+(3cdot21cdotp 5^2)+5^3$$



                      The binomial coefficients count the ways to select elements from a set.



                      In your case, that is positions to place the vowels in the word.



                      There are $3cdot21^2cdotp 5$ ways to select 1 vowels and two consonants where the vowel is in the first, second, or third place. And so on.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Refer to Binomial Expansion



                        $$(21+5)^3=21^3+tbinom 31cdotp 21^2cdotp5+tbinom 32cdotp21cdotp 5^2+5^3$$



                        so$$26^3-21^3= (3cdot21^2cdotp 5)+(3cdot21cdotp 5^2)+5^3$$



                        The binomial coefficients count the ways to select elements from a set.



                        In your case, that is positions to place the vowels in the word.



                        There are $3cdot21^2cdotp 5$ ways to select 1 vowels and two consonants where the vowel is in the first, second, or third place. And so on.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Refer to Binomial Expansion



                          $$(21+5)^3=21^3+tbinom 31cdotp 21^2cdotp5+tbinom 32cdotp21cdotp 5^2+5^3$$



                          so$$26^3-21^3= (3cdot21^2cdotp 5)+(3cdot21cdotp 5^2)+5^3$$



                          The binomial coefficients count the ways to select elements from a set.



                          In your case, that is positions to place the vowels in the word.



                          There are $3cdot21^2cdotp 5$ ways to select 1 vowels and two consonants where the vowel is in the first, second, or third place. And so on.






                          share|cite|improve this answer









                          $endgroup$



                          Refer to Binomial Expansion



                          $$(21+5)^3=21^3+tbinom 31cdotp 21^2cdotp5+tbinom 32cdotp21cdotp 5^2+5^3$$



                          so$$26^3-21^3= (3cdot21^2cdotp 5)+(3cdot21cdotp 5^2)+5^3$$



                          The binomial coefficients count the ways to select elements from a set.



                          In your case, that is positions to place the vowels in the word.



                          There are $3cdot21^2cdotp 5$ ways to select 1 vowels and two consonants where the vowel is in the first, second, or third place. And so on.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          Graham KempGraham Kemp

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