Emma’s Dilemma

There are a total of 24 permutations of a 4-letter word with none of the letters the same. This could be written as 4*3*2*1. It is written like this because the result is 4 times greater than the result for 3 letters, all different. There are also 4 letters in the word.

I shall investigate now, 4 letters with 2 letters the same. This time the H has been substituted for an A

. FAAK

2. FKAA

3. FAKA

4. AFKA

5. AFAK

6. AAFK

7. AKAF

8. AAKF

9. AKFA

0. KFAA

1. KAAF

2. KAFA

There are a total of 12 permutations for using a 4-lettered word with 2 of the letters the same. This is again the half of the original amount if 2 of the letters are the same and is written like this as there are 4 letters with 2 of them the same and 2 the same. (4*3*2*1)/(2*1)

4 letters, 3 the same.

. AAAK

2. AAKA

3. AKAA

4. KAAA

There are 4 possible permutations of this which is 24/6, which is also the original value, divided by 6. or (4*3*2*1)/(3*2*1). It is written this way, as there are 4 letters with 3 of them the same.

Table 1

In this table there are all of the results from the 3 and 4 lettered words with none of the letters repeated, some of the results are predicted using the pattern. At the bottom of the table is the formula where n is the number of letters in the word.

Number of letters in word

Number of permutations

How to find total permutations

3

6

3*2*1

4

24

4*3*2*1

5

20

5*4*3*2*1

6

720

6*5*.....*1

7

5040

7*6*.....*1

8

40320

8*7.....*1

9

362880

9*8...*1

0

3628800

0*9...*1

n

n*(n-1)*(n-2)...*1

This is the formula in use with a 4-lettered word, as we know the results for it. 4*(4-1)*(4-2)*(4-3) which gives 24. This proves that the formula is correct and will work for any word if none of the letters are the same.

This table shows words with 2 or 3 letters the same; the results for words with 5-10 letters have been predicted using the above formula.

Table 2

Number of letters

Permutations for 2 letters the same

Formula

Permutations for 3 letters the same

Formula

3

3

3*2*1/(2*1)

3*2*1/(3*2*1)

4

2

4*3*2*1/(2*1)

4

4*3...*1/(3*2*1)

5

60

5*4...*1/(2*1)

20

5*3...*1/(3*2*1)

6

360

6*5.... *1/(2*1)

20

6*5.... *1/(3*2*1)

7

2520

7*6.... *1/(2*1)

840

7*6.... *1/(3*2*1)

8

40320

8*7.... *1/(2*1)

6720

8*7.... *1/(3*2*1)

9

81440

9*8.... *1/(2*1)

60480

9*8....*1/(3*2*1)

0

814400

0*9....*1/(2*1)

604800

0*9....*1/(3*2*1)

n

n*(n-1)*(n-2)...*1/(2*1)

n*(n-1)*(n-2)...*1/(3*2*1)

The table shows that if there are 2 letters the same then the formula ends in 2*1. If there are 3 letters the same then the formula ends in 3*2*1. So the formula for a letter being used more than once is

n*(n-1)*(n-2)*(n-3)....*1/m*(m-1)*(m-2)*(m-3)...*1. In this formula m is how many times a letter is used. This is how the formula works: if you have a 5 lettered word with 3 letters the same: 5*(5-1)*(5-2)*(5-3)*(5-4)/3*(3-1)*(3-2) which is 20. To prove this here are all of the possible permutations for a 5-lettered word, 2 the same.

. GAGOG

2. GAGGO

3. GAOGG

4. GOAGG

5. GOGAG

6. GOGGA

7. GGOGA

8. GGOAG

9. GGAOG

0. GGAGO

1. GGGAO

2. GGGOA

3. AGOGG

4. AGGOG

5. AGGGO

6. AOGGG

7. OGAGG

8. OGGAG

9. OGGGA

20. OAGGG

This proves that the formula is correct for words with a letter repeated.

This formula will only work if there is one set of letters the same; if there are more then a different formula is needed.

This time I will look at a 4-lettered word consisting of 2 letters the same and 2 letters the same.

The 4-lettered word that will be used is AABB.

. AABB

2. ABBA

3. ABAB

4. BBAA

5. BABA

6. BAAB

There are a total of 6 permutations for this word, which could be written as (4*3*2*1) /((2*1)*(2*1)) as there are 4 letters in the word and there are 2 letters repeated twice. The first bracket represents 1 set of repeated letters and the second brackets represents the second set of letters. I predict that if there are 5 letters with 3 the same and 2 the same then there will be 10 permutations. This is using the formula that the total number of letters in the word is always divided by the amount of repeated letters.

5 letters, 3 same, 2 same

. AAABB

2. AABBA

3. AABAB

4. ABAAB

5. ABABA

6. ABBAA

7. BBAAA

8. BAABA

9. BAAAB

0. BABAA

This could be written as 5*4*3*2*1/((3*2*1)*(2*1)) as there are 5 letters with three the same and 2 the same. The first bracket represents 1 set of repeated letters and the second brackets represents the second set of letters.

I shall now investigate that if you have a 5-lettered word with 2 sets of repeated letters the same this is the result.

. AABBC

2. AABCB

3. AACBB

4. ABBCA

5. ABBAC

6. ABAAC

7. ABACA

8. ABCAA

9. ACBBA

0. ACABB

1. ACBAB

2. BBAAC

3. BBACA

4. BBCAA

5. BACAB

6. BACBA

7. BABCA

8. BABAC

9. BAACB

20. BAABC

21. BCAAB

22. BCABA

23. BCBAA

24. CAABB

25. CABBA

26. CABAB

27. CBAAB

28. CBBAA

29. CBAAB

30. CBABA

There are a total of 30 permutations for a 5 letter word with 2 the same, 2 the same and 1 different. This could be written as 5*4*3*2*1/((2*1)*(2*1)) as there are 5 letters in the word and there are 2 lots of the same letter. The first bracket represents 1 set of repeated letters and the second brackets represents the second set of letters

Table 3

Number of letters

Permutations for 2 letters the same, 2 the same

Formula

4

6

4*3...*1/((2*1)*(2*1))

5

30

5*3...*1/((2*1)*(2*1))

6

80

6*5...*1/((2*1)*(2*1))

7

260

7*6...*1/((2*1)*(2*1))

8

0080

8*7...*1/((2*1)*(2*1))

9

90720

9*8...*1/((2*1)*(2*1))

0

9070200

0*9...*1/((2*1)*(2*1))

n

n*(n-1)*(n-2)...*1/((2*1)*(2*1))

This time I shall investigate words, which have 2 sets of 3 letters the same.

. AAABBB

2. AABBBA

3. AABBAB

4. AABABB

5. ABBBAA

6. ABBAAB

7. ABBABA

8. ABABAB

9. ABABBA

0. ABAABB

1. BBBAAA

2. BBAAAB

3. BBAABA

4. BBABAA

5. BABBAA

6. BAABBA

7. BAABAB

8. BABABA

9. BABBAA

20. BABAAB

There are 20 permutations for a word, which has 3 letters the same and 3 the same. This could be written as 6*5*4*3*2*1/((3*2*1)*(3*2*1)) as there are 6 letters in the word and there are 2 lots of 3 letters which are the same. The first bracket is the first set of repeated letters and the second bracket is the second set of repeated letters.

Using the pattern from previous formulas and the rule for if there are 2 letters the same and 2 the same I believe that the next set of results for 7 letters will be 140

(7*6...*1/((3*2*1)*(3*2*1)) permutations if there are 3 letters the same and 3 the same. I think this because table 2 that if letters are repeated and there is one that isn't, this letter is not included in the dividing part of the formula.

Table 4

Number of letters

Permutations for 3 letters the same and 3 the same

Formula

6

20

6*5...*1/((3*2*1) (3*2*1))

7

40

7*6...*1/((3*2*1) (3*2*1))

8

120

8*7...*1/((3*2*1) (3*2*1))

9

0080

9*8...*1/((3*2*1) (3*2*1))

0

00800

0*9...*1/((3*2*1) (3*2*1))

n

n(n-1)*(n-2)...*1/((3*2*1)*(3*2*1))

In the results I have noticed that if there is one set of repeated letters then there is only 1 set of brackets under the division sign. If there are 2 sets of repeated letters then there are 2 sets of brackets. So using this pattern if there were 3 sets of repeated letters there would be 3 sets of brackets, and if there were 7 sets of repeated letters then there would be 7 brackets. So this gives the formula

n*(n-1)*(n-2...*1/((m*(m-1)*(m-2)...*1)*(p*(p-1)*(p-2))...*1 where p is the second set of repeated letters. For example if there was a 7-lettered word with 2 letters the same and 3 letters the same then the formula would be 7*6*5...*1/((2*1)*(3*2*1)) which is 420 permutations.

To show this I am going to use the word MATHEMATICS. In this word there are 3 sets of repeated letters so there will be 3 divisions. This means that the formula will be 11*10*9...*1/((2*1)*(2*1)*(2*1)). This means that there are a possible 4989600 permutations for an 11-lettered word with 3 sets of repeated letters.

The conclusion to this investigation is that I have found that however many sets of repeated letters there are in a word gives the amount of brackets in the formula. For example if there are 3 sets of repeated letters then there are 3 brackets. If there are 10 sets of repeated letters there will be 10 sets of brackets etc. if the word has 15 letters i.e. 5 the same and 10 the same then the formula still has the 15*14*13...*1, which is 1307674368000. This is then divided by 5*4...*1, then by 10*9...*1 which gives a total of 3003 permutations.

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