EMMA'S DILEMMA

Let’s work out the formula:

   

1 letter: 1

2 letters: 1*2

3 letters: 1*2*3

4 letters: 1*2*3*4

And so on.

So the formula is N! =A

N= Number of letters

! = Number of arrangements for (N-1)

A= Number of arrangements

I worked out the formula, and now lets see if the formula works by doing another table and matching with the prediction table.  

Accurate table:

Table 2

 

The prediction table matches with the accurate table. So the formula works. But lets see if it will work on a 5-lettered word.

5-lettered Word:

   I deleted all the arrangements to have some space.

   

There are 120 arrangements in a 5-lettered word. Lets use the formula and see if it is correct.

5*24=120

The formula works.

So to find the total arrangements of any letter, you have to start by multiplying by 1 to the number of letters of that word. For example, to find the total arrangement of a 6 lettered word, you have to do this: 1*2*3*4*5*6=720 arrangements. This is called factorial. Factorial is known by (!) which can be found on a scientific calculator. So the formula to find the total arrangement of a word with no repeated letters is N! =A   

N= Number of letters

! = Number of arrangements for (N-1)

A= Number of arrangements

Part 2

I will start part 2, by finding the number of arrangements of the word EMMA and other words that has repeated letters.

1. EMMA   7.   MEMA
2. EMAM   8.   MAME
3. EAMM   9.   MMAE
4. AMME   10. MMEA
5. AMEM   11. MAEM
6. AEMM   12. MEAM

I have found 12 arrangements in the word EMMA. Now lets see how many arrangements does a 3-lettered word have with 2 same letters.

3-lettered word with 2 same letters:

1. CNN
2. NNC
3. NCN

I have found only 3 arrangements in a 3-lettered word with 2 same letters.

I have realised that the word LUCY has more arrangements then EMMA. EMMA has 12 arrangements and LUCY has 24. EMMA’S arrangement is half of LUCY’S arrangement. In a 3-lettered word with 2 same letters I found 3 arrangements. I have realised that there is a pattern, because if a letter is repeated twice then the arrangement will be half of a no repeated letter. So if I want to find the formula, then I have to divide it by the number of repeated letters. If I don’t do it like this then the formula would not work and I will come with a wrong arrangement.

So I think the formula is N! /R =A

A= Number of arrangements

N= Number of letters

R= Number of repeated letters

!  = Number of arrangements for (N-1)

Lets see if the formula works with the word EMMA.

1*2*3*4=24/1*2=12

The formula works.

Lets try and work out the arrangement for a 4-lettered word with three same letters.

4-lettered word with 3 same letters:

1. ABBB
2. BABB
3. BBAB
4. BBBA

There are 4 arrangements in a 4-lettered word with 3 same letters.

Lets use the formula that I have created and lets see if it works.

1*2*3*4=24/3=8

My formula does not work, but it will work on a word with 2 letters repeated.

Lets see what’s wrong with my formula.    

2 letters are same, then:

3 letters: 1*2*3/1*2

4 letters: 1*2*3*4/1*2

5 letters: 1*2*3*4*5/1*2

So to get the arrangement of a word with two same letters, then you have to divide it by 2. Lets see how you get the arrangement of a word with 3 letters same.

 

3 letters are same, then:

3 letters: 1*2*3/1*2*3

4 letters: 1*2*3*4/1*2*3

5 letters: 1*2*3*4*5/1*2*3

And so on.

I have realised that, to find the arrangement of a 4 lettered word with 3 same letters.

I have to get the arrangement of the 4-lettered word, which is 24. Then I have to find the arrangement of the repeated letters. So there are 3 letters same in a 4 lettered word. So the arrangement of the repeated letter is 6. Because the arrangement of a 3 lettered word is 6.

Then you have to divide 24 by 6 and you will get the answer, which is 4.

My prediction is that to find the arrangement of a word with repeated letters, you have to first get the arrangement of the word. Then you have get the number of arrangements of how many repeated letters there are. Then you divide both the number of arrangement of the word and the number of arrangement of the repeated letters. Then you will get the arrangement. This is my prediction.  

I have given a table below to show my prediction. I will call the prediction table, table 3

Prediction table:

3 letters same:

Table 3

This is my prediction. Now lets work out the formula and check if the formula works.

If 2 letters are same, then:

3 letters: 1*2*3/1*2

4 letters: 1*2*3*4/1*2

5 letters: 1*2*3*4*5/1*2

And so on.

If 3 letters are same, then:

3 letters: 1*2*3/1*2*3

4 letters: 1*2*3*4/1*2*3

5 letters: 1*2*3*4*5/1*2*3

And so on.

So the formula is:

                          A= N! / R!

A= Number of arrangements

N= Number of letters

R= Number of repeated letters

!  = Number of arrangements for (N-1)

I worked out the formula, and now lets see if the formula works by doing another table and matching with the prediction table. I will call the accurate table, table 4. If table 4 matches with table 3, then the formula works.  

Accurate table:

3 letters same:

Table 4

The Accurate table matches with the Prediction table. So the formula works. But lets see if it will work on a 5-lettered word with 3 same letters.

5-lettered word with 3 same letters:

1. DADDY   11.   DYDDA
2. DADYD   12.   DYDAD
3. DAYDD   13.   AYDDD
4. DDADY   14.   ADYDD
5. DDAYD   15.   ADDYD
6. DDYAD    16.   ADDDY
7. DDYDA   17.   YDDDA
8. DDDAY   18.   YDADD
9. DDDYA   19.   YDDAD
10. DYADD   20.   YADDD

There are 20 arrangements in a 5-lettered word with 3 same letters.

Let’s use the formula and check if it is correct.

1*2*3*4*5 / 1*2*3 = 20

The formula works.

Now lets investigate the formula, and improve it:

N represents the number of letters same

R represents the divided number in the formula

From the formulas that I have found, I have noticed that the number N! is divided by the result of doing factorial with the number of repeated letters. For example, if you want to find the total arrangement of a 6 lettered word, which has 2 repeated letters, you will first do 1*2*3*4*5*6, which equals to 720 arrangements, and then do 1*2 (2 is the number of repeated letters) which is 2.then you will divide 720 by 2, that equals to 360. This can be made simpler like this:                    

6! / 2!

= (1*2*3*4*5*6) / (1*2)

= 720 / 2

= 360

If you don’t do it like this then the arrangement will be wrong. So there formula is:

A= N! / R!

A= Number of arrangements

N= Number of letters

R= Number of repeated letters

!  = Number of arrangements for (N-1)

Part 3

In part 3, I will investigate the number of different arrangements of various groups of letter. Like in a 4-lettered word, 1 letter is repeated twice and the other letter is also repeated twice.

I will work out the arrangement for a 4-lettered word which 1 letter is repeated twice and the other letter is also repeated twice.

4-lettered word:

      1.   AABB   3.   ABAB   5.   BABA

2. ABBA   4.   BBAA   6.   BAAB

There are 6 arrangements in a 4-lettered word which 1 letter is repeated twice and the other letter is also repeated twice.

Lets use the formula in part 2 and let’s see if the formula still works.

1*2*3*4

   1*2    = 12

So the formula does not work.

But f I

Now lets try a 6-lettered word, which 1 letter is repeated 3 times and the letter is also repeated 3 times:

     1.   AAABBB   11.   BBBAAA

     2.   AABABB   12.   BBABAA

3. AABBAB   13.   BBAABA
4. AABBBA   14.   BBAAAB
5. ABABBA   15.   BABAAB
6. ABABAB   16.   BABABA
7. ABAABB   17.   BABBAA
8. ABBAAB   18.   BAABBA
9. ABBABA   19.   BAABAB
10. ABBBAA   20.   BAAABB

There are 20 arrangements.

There is a pattern. The formula will still work if I double the repeated letters arrangement.

For example for a 4-lettered word which 1 letter is repeated twice and the other letter is also repeated twice.

If I do the formula it will be like this:

1*2*3*4

        (1*2*1*2)    =6

The formula works.

To use the formula for a 6 lettered word, which 1 letter is repeated twice and the other letter is also repeated twice and the final letter is also repeated twice. To find the arrangement, if I triple the repeated letters arrangement. Then I will get the arrangement.

For example

1*2*3*4*5*6

             (1*2*1*2*1*2)      =90

I should get this arrangement.

Lets work out the arrangement, but not using the formula.

6-lettered word with 1 letter repeated twice and the second letter is also repeated twice and the final letter is also repeated twice:

1. AABBCC   31. BBAACC   61. CCAABB
2. AABCBC   32. BBACAC   62. CCABAB
3. AABCCB   33. BBACCA   63. CCABBA
4. AACBBC   34. BBCAAC   64. CCBAAB
5. AACBCB   35. BBCACA   65. CCBABA
6. AACCBB   36. BBCCAA   66. CCBBAA
7. ABABCC   37. BABACC   67. CACABB
8. ABACBC   38. BABCBC   68. CACBCB
9. ABACCB   39. BABCCB   69. CACBBC
10. ABBACC   40. BAABCC   70. CAACBB
11. ABBCAC   41. BAACBC   71. CAABCB
12. ABBCCA   42. BAACCB   72. CAABBC
13. ABCABC   43. BACBAC   73. CABCAB
14. ABCACB   44. BACBCA   74. CABCBA
15. ABCBAC   45. BACABC   75. CABACB
16. ABCBCA   46. BACACB   76. CABABC
17. ABCCAB   47. BACCBA   77. CABBCA
18. ABCCBA   48. BACCAB   78. CABBAC
19. ACABBC   49. BCBAAC   79. CBCAAB
20. ACABCB   50. BCBACA   80. CBCABA
21. ACACBB   51. BCBCAA   81. CBCBAA
22. ACBCBA   52. BCACAB   82. CBABCA
23. ACBCAB   53. BCACBA   83. CBABCA
24. ACBBAC   54. BCAABC   84. CBAACB
25. ACBBCA   55. BCAACB   85. CBAABC
26. ACBABC   56. BCABAC   86. CBACAB
27. ACBACB   57. BCABCA   87. CBACBA
28. ACCABB   58. BCCBAA   88. CBBCAA
29. ACCBAB   59. BCCABA   89. CBBACA
30. ACCBBA   60. BCCAAB   90. CBBAAC

I have found 90 arrangements. This arrangement also matches with the formula. So the formula worked and gave me the right arrangement.

What if a word had three letters same and the other 2 was also same. Lets work out the arrangement and find the right arrangement with the formula.

5-lettered word with 1 letter repeating 3 times and the other letter repeating twice:

1. AAABB   6.   ABBAA
2. AABAB   7.   BAAAB
3. AABBA   8.   BAABA
4. ABAAB   9.   BABAA
5. ABABA   10. BBAAA

I have found 10 arrangements. The formula for this will be like this, because there aren’t any changes to be made because you just have to multiply the repeated letters.

1*2*3*4*5

                                                           (1*2*3*1*2)     =10

For example: 3*2*1 is one and 1*2 is another one. If you calculate the first one, which is 3*2*1=6 and multiply it by 2 you will get 12. I got number 2 from the second one; which 1*2=2. Then I multiplied it by 6. When I multiplied it gave me number 12. With that number I divided it by the number of letter arrangements. There it gave me the right arrangements.

 So the formula worked.

To use the formula, I can find out the total arrangements of any letter. Repeated or no repeated.

 

 From this I have worked out the formula to find out the number of different arrangements.

So the formula is:

A=N! / R!

     

A= Number of arrangements

N= Number of letters

R= Number of repeated letters

!  = Number of arrangements for (N-1)

This formula can also be done in a calculator. By typing in the number of letters and then pressing '!’. You should get the arrangement.

These letters I put are important. Without these letters you cannot find the arrangement. The letter ‘A’ tells you what the arrangement is. Without that letter you would not know the arrangement. So the ‘A’ letter is important. The letter ‘N’ tells you how many letters are in the word. With this letter you need to find out the arrangement of a no repeated word. So this letter is important. The Letter ‘R’ is the final letter in the formula. This letter tells you how many repeated letters there are. Without this letter you cannot find the arrangement of a word with repeated letters in it. Last of all is the icon ‘!’. This icon is very important. This icon is used for everything. To find an arrangement for a repeated letter or no repeated letter. This icon cannot be removed from the formula because without that icon you cannot find the formula.    

EVALUATION

From this investigation I have learned how to find arrangements and work out the formula. This investigation was quiet exciting but it was also quite tiring because of finding the arrangement. This investigation was long and quite hard as well. It was quite challenging investigation.