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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?

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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asked 1 hour ago

StudentStudent

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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

share|cite|improve this question

asked 1 hour ago

StudentStudent

153

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.

$endgroup$

add a comment | 

$begingroup$

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

asked 1 hour ago

StudentStudent

153

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.

$endgroup$

share|cite|improve this question

asked 1 hour ago

StudentStudent

153

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.

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asked 1 hour ago

StudentStudent

153

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.

asked 1 hour ago

StudentStudent

153

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asked 1 hour ago

StudentStudent

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3 Answers
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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

edited 1 hour ago

answered 1 hour ago

David G. StorkDavid G. Stork

11.6k41534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get it. Thanks so much
  $endgroup$
  – Student
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

answered 1 hour ago

heropupheropup

64.9k764103

$endgroup$

add a comment | 

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

answered 1 hour ago

Graham KempGraham Kemp

87.6k43578

$endgroup$

add a comment | 

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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

edited 1 hour ago

answered 1 hour ago

David G. StorkDavid G. Stork

11.6k41534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get it. Thanks so much
  $endgroup$
  – Student
  1 hour ago
  

  
  
  
  

add a comment | 

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

edited 1 hour ago

answered 1 hour ago

David G. StorkDavid G. Stork

11.6k41534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get it. Thanks so much
  $endgroup$
  – Student
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 1 hour ago

answered 1 hour ago

David G. StorkDavid G. Stork

11.6k41534

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

David G. StorkDavid G. Stork

11.6k41534

answered 1 hour ago

David G. StorkDavid G. Stork

11.6k41534

answered 1 hour ago

David G. StorkDavid G. Stork

11.6k41534

1

$begingroup$

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

answered 1 hour ago

heropupheropup

64.9k764103

$endgroup$

add a comment | 

1

$begingroup$

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

answered 1 hour ago

heropupheropup

64.9k764103

$endgroup$

add a comment | 

$begingroup$

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

answered 1 hour ago

heropupheropup

64.9k764103

$endgroup$

share|cite|improve this answer

answered 1 hour ago

heropupheropup

64.9k764103

answered 1 hour ago

heropupheropup

64.9k764103

answered 1 hour ago

heropupheropup

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

answered 1 hour ago

Graham KempGraham Kemp

87.6k43578

$endgroup$

add a comment | 

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

answered 1 hour ago

Graham KempGraham Kemp

87.6k43578

$endgroup$

add a comment | 

$begingroup$

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

answered 1 hour ago

Graham KempGraham Kemp

87.6k43578

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Graham KempGraham Kemp

87.6k43578

answered 1 hour ago

Graham KempGraham Kemp

87.6k43578

answered 1 hour ago

Graham KempGraham Kemp

87.6k43578