Why do I get two different answers for this counting problem?Combinatorics question about english letters (with consonants and vowels)Permutation and CombinationsWords counting problemTwo different answers for combinatorics question - which is correct?Two different answers - cubes and colorsHow many Strings of 6 letters contain: Exactly one Vowel, At least one Vowel?Ways to arrange a word so that no vowel is isolated between two consonantsEx 2., Combinatorics, Harris - Eleven letter sequences from the 26-letter alphabet containing exactly three vowels -String of letters and ways to have at least one vowelHow many different arrangements using 5 letters of the word INTEGRAL, if no two vowels are adjacent?

If a centaur druid Wild Shapes into a Giant Elk, do their Charge features stack?

How would photo IDs work for shapeshifters?

Are white and non-white police officers equally likely to kill black suspects?

Is every set a filtered colimit of finite sets?

Check if two datetimes are between two others

Information to fellow intern about hiring?

Why was the "bread communication" in the arena of Catching Fire left out in the movie?

Is ipsum/ipsa/ipse a third person pronoun, or can it serve other functions?

Hosting Wordpress in a EC2 Load Balanced Instance

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

Finding files for which a command fails

Unbreakable Formation vs. Cry of the Carnarium

What do you call words made from common English words?

Piano - What is the notation for a double stop where both notes in the double stop are different lengths?

What is GPS' 19 year rollover and does it present a cybersecurity issue?

What does 'script /dev/null' do?

Doomsday-clock for my fantasy planet

Can I find out the caloric content of bread by dehydrating it?

Is Social Media Science Fiction?

Can a planet have a different gravitational pull depending on its location in orbit around its sun?

Is it wise to focus on putting odd beats on left when playing double bass drums?

Denied boarding due to overcrowding, Sparpreis ticket. What are my rights?

Is this food a bread or a loaf?

Does a dangling wire really electrocute me if I'm standing in water?



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


Combinatorics question about english letters (with consonants and vowels)Permutation and CombinationsWords counting problemTwo different answers for combinatorics question - which is correct?Two different answers - cubes and colorsHow many Strings of 6 letters contain: Exactly one Vowel, At least one Vowel?Ways to arrange a word so that no vowel is isolated between two consonantsEx 2., Combinatorics, Harris - Eleven letter sequences from the 26-letter alphabet containing exactly three vowels -String of letters and ways to have at least one vowelHow many different arrangements using 5 letters of the word INTEGRAL, if no two vowels are adjacent?













19












$begingroup$


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?










share|cite|improve this question









$endgroup$







  • 4




    $begingroup$
    Nearly right, you're only missing factors of C(3,1) and C(3,2) on the terms respectively. So that you handle the possibilities CCV, CVC, VCC and CVV, VCV, VVC.
    $endgroup$
    – smci
    Apr 4 at 20:36






  • 4




    $begingroup$
    Also, you could have intuited those terms by considering the binomial expansion $(21 + 5)^3$
    $endgroup$
    – smci
    Apr 4 at 20:37















19












$begingroup$


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?










share|cite|improve this question









$endgroup$







  • 4




    $begingroup$
    Nearly right, you're only missing factors of C(3,1) and C(3,2) on the terms respectively. So that you handle the possibilities CCV, CVC, VCC and CVV, VCV, VVC.
    $endgroup$
    – smci
    Apr 4 at 20:36






  • 4




    $begingroup$
    Also, you could have intuited those terms by considering the binomial expansion $(21 + 5)^3$
    $endgroup$
    – smci
    Apr 4 at 20:37













19












19








19


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?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 4 at 4:43







user660730














  • 4




    $begingroup$
    Nearly right, you're only missing factors of C(3,1) and C(3,2) on the terms respectively. So that you handle the possibilities CCV, CVC, VCC and CVV, VCV, VVC.
    $endgroup$
    – smci
    Apr 4 at 20:36






  • 4




    $begingroup$
    Also, you could have intuited those terms by considering the binomial expansion $(21 + 5)^3$
    $endgroup$
    – smci
    Apr 4 at 20:37












  • 4




    $begingroup$
    Nearly right, you're only missing factors of C(3,1) and C(3,2) on the terms respectively. So that you handle the possibilities CCV, CVC, VCC and CVV, VCV, VVC.
    $endgroup$
    – smci
    Apr 4 at 20:36






  • 4




    $begingroup$
    Also, you could have intuited those terms by considering the binomial expansion $(21 + 5)^3$
    $endgroup$
    – smci
    Apr 4 at 20:37







4




4




$begingroup$
Nearly right, you're only missing factors of C(3,1) and C(3,2) on the terms respectively. So that you handle the possibilities CCV, CVC, VCC and CVV, VCV, VVC.
$endgroup$
– smci
Apr 4 at 20:36




$begingroup$
Nearly right, you're only missing factors of C(3,1) and C(3,2) on the terms respectively. So that you handle the possibilities CCV, CVC, VCC and CVV, VCV, VVC.
$endgroup$
– smci
Apr 4 at 20:36




4




4




$begingroup$
Also, you could have intuited those terms by considering the binomial expansion $(21 + 5)^3$
$endgroup$
– smci
Apr 4 at 20:37




$begingroup$
Also, you could have intuited those terms by considering the binomial expansion $(21 + 5)^3$
$endgroup$
– smci
Apr 4 at 20:37










6 Answers
6






active

oldest

votes


















27












$begingroup$

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











$endgroup$












  • $begingroup$
    I get it. Thanks so much
    $endgroup$
    – user660730
    Apr 4 at 4:52






  • 2




    $begingroup$
    It's only missing factors of C(3,1) and C(3,2) on the terms respectively.
    $endgroup$
    – smci
    Apr 4 at 20:35


















59












$begingroup$

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









$endgroup$




















    33












    $begingroup$

    When wondering why (or if) something is wrong, a general approach is to make the numbers as small as you can while still maintaining the error but so that you can brute-force the answer. So let's try an alphabet with one vowel A and one consonant B. (Note that this way you can never prove that something is correct, but it's a valuable tool for finding some errors.)



    You get 1 word with no vowels (BBB), 3 words with one vowel (ABB, BAB, BBA), 3 words with two vowels (AAB, ABA, BAA), and 1 word with three vowels (AAA). Your first approach finds the correct number, $2^3-1^3=7$.



    However, the second approach claims for example that there is only $1 cdot 1^2 = 1$ word with exactly one vowel. From here, you can try to see why it gives the wrong answer work (which has already been answered by the other answers).






    share|cite|improve this answer









    $endgroup$




















      7












      $begingroup$

      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









      $endgroup$




















        2












        $begingroup$

        From a generating function perspective, if $V$ is the number of vowels, $C$ is the number of consonants, and $L$ is the total number of letters, then we have that the number of ways to have one vowel followed by two consonants is $VCC$, or $5*21*21$. We also have $CVC$ and $CCV$. Since multiplication is communative, we have that $VCC+CVC+CCV=3VCC$, so adding those together gives $3VCC$, or $3*5*21*21$. Similarly, for two vowels we have $3VVC=3*5*5*21$. For $VVV$, there's no other way to arrange the order, so it's just $1*5*5*5$.



        So your error was in calculating only $VCC$. That is, you calculated the number of ways of getting a particular order, and didn't take into account the different orders possible.



        We can also look at what happens when we take the total number of sequences, $L^3$. This is equal to $(C+V)^3$. That gives eight terms, $CCC, CCV, CVC, CVV, VCC, VCV, VVC, textand VVV$. Collecting terms that are equivalent up to commutation gives us the binomial formula results: $C^3+3C^2V+3CV^2+V^3$. Subtracting off $C^3$ gives $3C^2V+3CV^2+V^3$.






        share|cite|improve this answer









        $endgroup$




















          0












          $begingroup$

          In your second case $(5cdot21^2)+(5^2cdot21)+5^3=2855$, the first term represents selecting 1 vowel and any two constants. But this represents ABB, BAB, and BBA as one entity, so multiply it by 3 since the vowel could be in any position. Likewise the second term represents one constant and any combinations of vowels, but the constant can be in any of the three positions, so multiply that by 3 as well.



          $3(5cdot21^2)+3(5^2cdot21)+5^3=2855 = 8315$



          For the first term, you can represent vowel, constant, constant as $5cdot21cdot21$, constant, vowel, constant as $21cdot5cdot21$, and constant, constant, vowel as $21cdot21cdot5$, adding those together gives you $3(5cdot21^2)$






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3174227%2fwhy-do-i-get-two-different-answers-for-this-counting-problem%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown
























            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            27












            $begingroup$

            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











            $endgroup$












            • $begingroup$
              I get it. Thanks so much
              $endgroup$
              – user660730
              Apr 4 at 4:52






            • 2




              $begingroup$
              It's only missing factors of C(3,1) and C(3,2) on the terms respectively.
              $endgroup$
              – smci
              Apr 4 at 20:35















            27












            $begingroup$

            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











            $endgroup$












            • $begingroup$
              I get it. Thanks so much
              $endgroup$
              – user660730
              Apr 4 at 4:52






            • 2




              $begingroup$
              It's only missing factors of C(3,1) and C(3,2) on the terms respectively.
              $endgroup$
              – smci
              Apr 4 at 20:35













            27












            27








            27





            $begingroup$

            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 Apr 4 at 7:09

























            answered Apr 4 at 4:48









            David G. StorkDavid G. Stork

            12.1k41735




            12.1k41735











            • $begingroup$
              I get it. Thanks so much
              $endgroup$
              – user660730
              Apr 4 at 4:52






            • 2




              $begingroup$
              It's only missing factors of C(3,1) and C(3,2) on the terms respectively.
              $endgroup$
              – smci
              Apr 4 at 20:35
















            • $begingroup$
              I get it. Thanks so much
              $endgroup$
              – user660730
              Apr 4 at 4:52






            • 2




              $begingroup$
              It's only missing factors of C(3,1) and C(3,2) on the terms respectively.
              $endgroup$
              – smci
              Apr 4 at 20:35















            $begingroup$
            I get it. Thanks so much
            $endgroup$
            – user660730
            Apr 4 at 4:52




            $begingroup$
            I get it. Thanks so much
            $endgroup$
            – user660730
            Apr 4 at 4:52




            2




            2




            $begingroup$
            It's only missing factors of C(3,1) and C(3,2) on the terms respectively.
            $endgroup$
            – smci
            Apr 4 at 20:35




            $begingroup$
            It's only missing factors of C(3,1) and C(3,2) on the terms respectively.
            $endgroup$
            – smci
            Apr 4 at 20:35











            59












            $begingroup$

            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









            $endgroup$

















              59












              $begingroup$

              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









              $endgroup$















                59












                59








                59





                $begingroup$

                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 Apr 4 at 5:00









                heropupheropup

                65.4k865104




                65.4k865104





















                    33












                    $begingroup$

                    When wondering why (or if) something is wrong, a general approach is to make the numbers as small as you can while still maintaining the error but so that you can brute-force the answer. So let's try an alphabet with one vowel A and one consonant B. (Note that this way you can never prove that something is correct, but it's a valuable tool for finding some errors.)



                    You get 1 word with no vowels (BBB), 3 words with one vowel (ABB, BAB, BBA), 3 words with two vowels (AAB, ABA, BAA), and 1 word with three vowels (AAA). Your first approach finds the correct number, $2^3-1^3=7$.



                    However, the second approach claims for example that there is only $1 cdot 1^2 = 1$ word with exactly one vowel. From here, you can try to see why it gives the wrong answer work (which has already been answered by the other answers).






                    share|cite|improve this answer









                    $endgroup$

















                      33












                      $begingroup$

                      When wondering why (or if) something is wrong, a general approach is to make the numbers as small as you can while still maintaining the error but so that you can brute-force the answer. So let's try an alphabet with one vowel A and one consonant B. (Note that this way you can never prove that something is correct, but it's a valuable tool for finding some errors.)



                      You get 1 word with no vowels (BBB), 3 words with one vowel (ABB, BAB, BBA), 3 words with two vowels (AAB, ABA, BAA), and 1 word with three vowels (AAA). Your first approach finds the correct number, $2^3-1^3=7$.



                      However, the second approach claims for example that there is only $1 cdot 1^2 = 1$ word with exactly one vowel. From here, you can try to see why it gives the wrong answer work (which has already been answered by the other answers).






                      share|cite|improve this answer









                      $endgroup$















                        33












                        33








                        33





                        $begingroup$

                        When wondering why (or if) something is wrong, a general approach is to make the numbers as small as you can while still maintaining the error but so that you can brute-force the answer. So let's try an alphabet with one vowel A and one consonant B. (Note that this way you can never prove that something is correct, but it's a valuable tool for finding some errors.)



                        You get 1 word with no vowels (BBB), 3 words with one vowel (ABB, BAB, BBA), 3 words with two vowels (AAB, ABA, BAA), and 1 word with three vowels (AAA). Your first approach finds the correct number, $2^3-1^3=7$.



                        However, the second approach claims for example that there is only $1 cdot 1^2 = 1$ word with exactly one vowel. From here, you can try to see why it gives the wrong answer work (which has already been answered by the other answers).






                        share|cite|improve this answer









                        $endgroup$



                        When wondering why (or if) something is wrong, a general approach is to make the numbers as small as you can while still maintaining the error but so that you can brute-force the answer. So let's try an alphabet with one vowel A and one consonant B. (Note that this way you can never prove that something is correct, but it's a valuable tool for finding some errors.)



                        You get 1 word with no vowels (BBB), 3 words with one vowel (ABB, BAB, BBA), 3 words with two vowels (AAB, ABA, BAA), and 1 word with three vowels (AAA). Your first approach finds the correct number, $2^3-1^3=7$.



                        However, the second approach claims for example that there is only $1 cdot 1^2 = 1$ word with exactly one vowel. From here, you can try to see why it gives the wrong answer work (which has already been answered by the other answers).







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Apr 4 at 13:47









                        JiKJiK

                        5,2231333




                        5,2231333





















                            7












                            $begingroup$

                            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









                            $endgroup$

















                              7












                              $begingroup$

                              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









                              $endgroup$















                                7












                                7








                                7





                                $begingroup$

                                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 Apr 4 at 5:04









                                Graham KempGraham Kemp

                                87.8k43578




                                87.8k43578





















                                    2












                                    $begingroup$

                                    From a generating function perspective, if $V$ is the number of vowels, $C$ is the number of consonants, and $L$ is the total number of letters, then we have that the number of ways to have one vowel followed by two consonants is $VCC$, or $5*21*21$. We also have $CVC$ and $CCV$. Since multiplication is communative, we have that $VCC+CVC+CCV=3VCC$, so adding those together gives $3VCC$, or $3*5*21*21$. Similarly, for two vowels we have $3VVC=3*5*5*21$. For $VVV$, there's no other way to arrange the order, so it's just $1*5*5*5$.



                                    So your error was in calculating only $VCC$. That is, you calculated the number of ways of getting a particular order, and didn't take into account the different orders possible.



                                    We can also look at what happens when we take the total number of sequences, $L^3$. This is equal to $(C+V)^3$. That gives eight terms, $CCC, CCV, CVC, CVV, VCC, VCV, VVC, textand VVV$. Collecting terms that are equivalent up to commutation gives us the binomial formula results: $C^3+3C^2V+3CV^2+V^3$. Subtracting off $C^3$ gives $3C^2V+3CV^2+V^3$.






                                    share|cite|improve this answer









                                    $endgroup$

















                                      2












                                      $begingroup$

                                      From a generating function perspective, if $V$ is the number of vowels, $C$ is the number of consonants, and $L$ is the total number of letters, then we have that the number of ways to have one vowel followed by two consonants is $VCC$, or $5*21*21$. We also have $CVC$ and $CCV$. Since multiplication is communative, we have that $VCC+CVC+CCV=3VCC$, so adding those together gives $3VCC$, or $3*5*21*21$. Similarly, for two vowels we have $3VVC=3*5*5*21$. For $VVV$, there's no other way to arrange the order, so it's just $1*5*5*5$.



                                      So your error was in calculating only $VCC$. That is, you calculated the number of ways of getting a particular order, and didn't take into account the different orders possible.



                                      We can also look at what happens when we take the total number of sequences, $L^3$. This is equal to $(C+V)^3$. That gives eight terms, $CCC, CCV, CVC, CVV, VCC, VCV, VVC, textand VVV$. Collecting terms that are equivalent up to commutation gives us the binomial formula results: $C^3+3C^2V+3CV^2+V^3$. Subtracting off $C^3$ gives $3C^2V+3CV^2+V^3$.






                                      share|cite|improve this answer









                                      $endgroup$















                                        2












                                        2








                                        2





                                        $begingroup$

                                        From a generating function perspective, if $V$ is the number of vowels, $C$ is the number of consonants, and $L$ is the total number of letters, then we have that the number of ways to have one vowel followed by two consonants is $VCC$, or $5*21*21$. We also have $CVC$ and $CCV$. Since multiplication is communative, we have that $VCC+CVC+CCV=3VCC$, so adding those together gives $3VCC$, or $3*5*21*21$. Similarly, for two vowels we have $3VVC=3*5*5*21$. For $VVV$, there's no other way to arrange the order, so it's just $1*5*5*5$.



                                        So your error was in calculating only $VCC$. That is, you calculated the number of ways of getting a particular order, and didn't take into account the different orders possible.



                                        We can also look at what happens when we take the total number of sequences, $L^3$. This is equal to $(C+V)^3$. That gives eight terms, $CCC, CCV, CVC, CVV, VCC, VCV, VVC, textand VVV$. Collecting terms that are equivalent up to commutation gives us the binomial formula results: $C^3+3C^2V+3CV^2+V^3$. Subtracting off $C^3$ gives $3C^2V+3CV^2+V^3$.






                                        share|cite|improve this answer









                                        $endgroup$



                                        From a generating function perspective, if $V$ is the number of vowels, $C$ is the number of consonants, and $L$ is the total number of letters, then we have that the number of ways to have one vowel followed by two consonants is $VCC$, or $5*21*21$. We also have $CVC$ and $CCV$. Since multiplication is communative, we have that $VCC+CVC+CCV=3VCC$, so adding those together gives $3VCC$, or $3*5*21*21$. Similarly, for two vowels we have $3VVC=3*5*5*21$. For $VVV$, there's no other way to arrange the order, so it's just $1*5*5*5$.



                                        So your error was in calculating only $VCC$. That is, you calculated the number of ways of getting a particular order, and didn't take into account the different orders possible.



                                        We can also look at what happens when we take the total number of sequences, $L^3$. This is equal to $(C+V)^3$. That gives eight terms, $CCC, CCV, CVC, CVV, VCC, VCV, VVC, textand VVV$. Collecting terms that are equivalent up to commutation gives us the binomial formula results: $C^3+3C^2V+3CV^2+V^3$. Subtracting off $C^3$ gives $3C^2V+3CV^2+V^3$.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Apr 4 at 16:46









                                        AcccumulationAcccumulation

                                        7,3052619




                                        7,3052619





















                                            0












                                            $begingroup$

                                            In your second case $(5cdot21^2)+(5^2cdot21)+5^3=2855$, the first term represents selecting 1 vowel and any two constants. But this represents ABB, BAB, and BBA as one entity, so multiply it by 3 since the vowel could be in any position. Likewise the second term represents one constant and any combinations of vowels, but the constant can be in any of the three positions, so multiply that by 3 as well.



                                            $3(5cdot21^2)+3(5^2cdot21)+5^3=2855 = 8315$



                                            For the first term, you can represent vowel, constant, constant as $5cdot21cdot21$, constant, vowel, constant as $21cdot5cdot21$, and constant, constant, vowel as $21cdot21cdot5$, adding those together gives you $3(5cdot21^2)$






                                            share|cite|improve this answer









                                            $endgroup$

















                                              0












                                              $begingroup$

                                              In your second case $(5cdot21^2)+(5^2cdot21)+5^3=2855$, the first term represents selecting 1 vowel and any two constants. But this represents ABB, BAB, and BBA as one entity, so multiply it by 3 since the vowel could be in any position. Likewise the second term represents one constant and any combinations of vowels, but the constant can be in any of the three positions, so multiply that by 3 as well.



                                              $3(5cdot21^2)+3(5^2cdot21)+5^3=2855 = 8315$



                                              For the first term, you can represent vowel, constant, constant as $5cdot21cdot21$, constant, vowel, constant as $21cdot5cdot21$, and constant, constant, vowel as $21cdot21cdot5$, adding those together gives you $3(5cdot21^2)$






                                              share|cite|improve this answer









                                              $endgroup$















                                                0












                                                0








                                                0





                                                $begingroup$

                                                In your second case $(5cdot21^2)+(5^2cdot21)+5^3=2855$, the first term represents selecting 1 vowel and any two constants. But this represents ABB, BAB, and BBA as one entity, so multiply it by 3 since the vowel could be in any position. Likewise the second term represents one constant and any combinations of vowels, but the constant can be in any of the three positions, so multiply that by 3 as well.



                                                $3(5cdot21^2)+3(5^2cdot21)+5^3=2855 = 8315$



                                                For the first term, you can represent vowel, constant, constant as $5cdot21cdot21$, constant, vowel, constant as $21cdot5cdot21$, and constant, constant, vowel as $21cdot21cdot5$, adding those together gives you $3(5cdot21^2)$






                                                share|cite|improve this answer









                                                $endgroup$



                                                In your second case $(5cdot21^2)+(5^2cdot21)+5^3=2855$, the first term represents selecting 1 vowel and any two constants. But this represents ABB, BAB, and BBA as one entity, so multiply it by 3 since the vowel could be in any position. Likewise the second term represents one constant and any combinations of vowels, but the constant can be in any of the three positions, so multiply that by 3 as well.



                                                $3(5cdot21^2)+3(5^2cdot21)+5^3=2855 = 8315$



                                                For the first term, you can represent vowel, constant, constant as $5cdot21cdot21$, constant, vowel, constant as $21cdot5cdot21$, and constant, constant, vowel as $21cdot21cdot5$, adding those together gives you $3(5cdot21^2)$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Apr 5 at 22:12









                                                Jason GoemaatJason Goemaat

                                                1504




                                                1504



























                                                    draft saved

                                                    draft discarded
















































                                                    Thanks for contributing an answer to Mathematics Stack Exchange!


                                                    • Please be sure to answer the question. Provide details and share your research!

                                                    But avoid


                                                    • Asking for help, clarification, or responding to other answers.

                                                    • Making statements based on opinion; back them up with references or personal experience.

                                                    Use MathJax to format equations. MathJax reference.


                                                    To learn more, see our tips on writing great answers.




                                                    draft saved


                                                    draft discarded














                                                    StackExchange.ready(
                                                    function ()
                                                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3174227%2fwhy-do-i-get-two-different-answers-for-this-counting-problem%23new-answer', 'question_page');

                                                    );

                                                    Post as a guest















                                                    Required, but never shown





















































                                                    Required, but never shown














                                                    Required, but never shown












                                                    Required, but never shown







                                                    Required, but never shown

































                                                    Required, but never shown














                                                    Required, but never shown












                                                    Required, but never shown







                                                    Required, but never shown







                                                    Popular posts from this blog

                                                    រឿង រ៉ូមេអូ និង ហ្ស៊ុយលីយេ សង្ខេបរឿង តួអង្គ បញ្ជីណែនាំ

                                                    Crop image to path created in TikZ? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Crop an inserted image?TikZ pictures does not appear in posterImage behind and beyond crop marks?Tikz picture as large as possible on A4 PageTransparency vs image compression dilemmaHow to crop background from image automatically?Image does not cropTikzexternal capturing crop marks when externalizing pgfplots?How to include image path that contains a dollar signCrop image with left size given

                                                    Romeo and Juliet ContentsCharactersSynopsisSourcesDate and textThemes and motifsCriticism and interpretationLegacyScene by sceneSee alsoNotes and referencesSourcesExternal linksNavigation menu"Consumer Price Index (estimate) 1800–"10.2307/28710160037-3222287101610.1093/res/II.5.31910.2307/45967845967810.2307/2869925286992510.1525/jams.1982.35.3.03a00050"Dada Masilo: South African dancer who breaks the rules"10.1093/res/os-XV.57.1610.2307/28680942868094"Sweet Sorrow: Mann-Korman's Romeo and Juliet Closes Sept. 5 at MN's Ordway"the original10.2307/45957745957710.1017/CCOL0521570476.009"Ram Leela box office collections hit massive Rs 100 crore, pulverises prediction"Archived"Broadway Revival of Romeo and Juliet, Starring Orlando Bloom and Condola Rashad, Will Close Dec. 8"Archived10.1075/jhp.7.1.04hon"Wherefore art thou, Romeo? To make us laugh at Navy Pier"the original10.1093/gmo/9781561592630.article.O006772"Ram-leela Review Roundup: Critics Hail Film as Best Adaptation of Romeo and Juliet"Archived10.2307/31946310047-77293194631"Romeo and Juliet get Twitter treatment""Juliet's Nurse by Lois Leveen""Romeo and Juliet: Orlando Bloom's Broadway Debut Released in Theaters for Valentine's Day"Archived"Romeo and Juliet Has No Balcony"10.1093/gmo/9781561592630.article.O00778110.2307/2867423286742310.1076/enst.82.2.115.959510.1080/00138380601042675"A plague o' both your houses: error in GCSE exam paper forces apology""Juliet of the Five O'Clock Shadow, and Other Wonders"10.2307/33912430027-4321339124310.2307/28487440038-7134284874410.2307/29123140149-661129123144728341M"Weekender Guide: Shakespeare on The Drive""balcony"UK public library membership"romeo"UK public library membership10.1017/CCOL9780521844291"Post-Zionist Critique on Israel and the Palestinians Part III: Popular Culture"10.2307/25379071533-86140377-919X2537907"Capulets and Montagues: UK exam board admit mixing names up in Romeo and Juliet paper"Istoria Novellamente Ritrovata di Due Nobili Amanti2027/mdp.390150822329610820-750X"GCSE exam error: Board accidentally rewrites Shakespeare"10.2307/29176390149-66112917639"Exam board apologises after error in English GCSE paper which confused characters in Shakespeare's Romeo and Juliet""From Mariotto and Ganozza to Romeo and Guilietta: Metamorphoses of a Renaissance Tale"10.2307/37323537323510.2307/2867455286745510.2307/28678912867891"10 Questions for Taylor Swift"10.2307/28680922868092"Haymarket Theatre""The Zeffirelli Way: Revealing Talk by Florentine Director""Michael Smuin: 1938-2007 / Prolific dance director had showy career"The Life and Art of Edwin BoothRomeo and JulietRomeo and JulietRomeo and JulietRomeo and JulietEasy Read Romeo and JulietRomeo and Julieteeecb12003684p(data)4099369-3n8211610759dbe00d-a9e2-41a3-b2c1-977dd692899302814385X313670221313670221