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To investigate the combination of arrangement of letters in Jeans name and then for her friend Emma.

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Introduction

Jean’s DilemmaPAGE 1

AIM:        To investigate the combination of arrangement  of letters in Jeans name and then for her friend Emma.

Plan:        First of all I will systematically write down the different combinations to Jeans name and then for her friend Emma.

Investigation:            Before I write the combinations down I will first make a prediction to the number of combinations. I predict there will be 24 combinations to Jeans name.

JEAN:

J  E  A  N          J  E  N  A          J  N  E  A          J  N  A  E          J  A  N  E          J  A  E  N          A  J  E  N          A  J  N  E          A  E  J  N          A  E  N  J          A  N  J  E          A  N  E  J          E  J  A  N          E  J  N  A          E  N  A  J          E  N  J  A          E  A  J  N          E  A  N  J          N  J  E  A          N  J  A  E            N  E  J  A          N  E  A  J          N  A  J  E          N  A  E  J

          There are 24 combinations, my prediction was correct, I will now try the combinations for Jeans friend, Emma, once again I predict there will be 24 combinations.

EMMA:

E  M  M  A          E  M  A  M          E  A  M  M         M  A  M  E         M  A  E  M          M  M  E  A          M  M  A  E          M  E  M  A         M  E  A  M         A  M  M  E           A  M  E  M          A  E  M  M

...read more.

Middle

JAMES: I predict there will be 100 combinations to the name James.

J  A  M  E  S           J  A  M  S  E          J  A  S  M  E         J  A  S  E  M             J  A  E  S  M           J  A  E  M  S          J  M  A  E  S         J  M  A  S  E               J  M  S  E  A           J  M  S  A  E          J  M  E  S  A         J  M  E  A  S                J  E  S  A  M           J  E  S  M  A          J  E  M  A  S         J  E  M  S  A               J  E  A  M  S           J  E  A  S  M          J  S  A  E  M         J  S  A  M  E                 J  S  M  A  E           J  S  M  E  A          J  S  E  A  M         J  S  E  M  A

          So far there are 24 combinations and I have only done one letter at the beginning, so I have a theory that if you substitute the J for any other letter that will give you another 24 combinations, so that’s would mean there are

                                                                                    PAGE 3

5 x 24 combinations = 120. If this theory is right then that would mean 4 letters would be 4 x the number of combinations of a 3 letter word. And then for 3 it would be 3 x the number of combinations of a 2 letter word, and then again the same for 2, 2 x the number of combinations of 1 letter. Which would mean 5 is, 5 x 4 x 3 x 2 x 1 = 120. And 4 x 3 x 2 x 1 = 24.

...read more.

Conclusion

#160; Y Z Z O Z     Y Z O Z Z     Y O Z Z Z

There are 20 combinations which is 120 / 3 / 2 / 1 / 1.

So

JAMES is 120 / 1 / 1 / 1 / 1 / 1.

SARAH is 120 / 2 / 1 / 1 / 1.

BILLI     is 120 / 2 / 2 / 1.

But…..

OZZZY  is 120 / 3 / 2 / 1 / 1 .

OZZZO is 120 / 3 / 2 / 2.

        Using this table I saw that when there is 3 letters you divide by 3 then 2, so maybe you actually divide by 1 as well but it makes no difference to the answer. So this would change the table slightly to

JAMES is 120 / 1 / 1 / 1 / 1 / 1.

SARAH is 120 / {2 / 1} / 1 / 1 / 1.

BILLI     is 120 / {2 / 1} / {2 / 1} /1.

OZZZY  is 120 / {3 / 2 / 1} / 1 / 1 .

OZZZO is 120 / {3 / 2 / 1} / {2 / 1}.

                                                                                PAGE 5

Using this whole theory you can work out any number of letters with any combination of the same letter, I will draw 2 charts to show this for 10 letters and 7 letters. To show 10x9x8x7x6x5x4x3x2x1 I will write 10! As shorthand. And the same for equations like 3/2/1 or 5/4/3/2/1.

10 letters

number of different letters

Same ones

Calculation

Combinations

10

All different

10!/1/1/1/1/1/1/1/1/1/1

3628800

9

1 1 1 1 1 1 1 1 2

10!/2!/1/1/1/1/1/1/1/1/1

1814400

8

1 1 1 1 1 1 2 2

10!/2!/2!/1/1/1/1/1/1

907200

7

1 1 1 1 2 2 2

10!/2!/2!/2!/1/1/1/1

453600

6

1 1 2 2 2 2

10!/2!/2!/2!/2!/1/1

226800

5

2 2 2 2 2

10!/2!/2!/2!/2!/2!

113400

4

3 3 2 2

10!/3!/3!/2!/2!

25200

3

3 3 3 1

10!/3!/3!/3!/1

16800

2

5 5

10!/5!/5!

252

7 letters

number of different letters

Same ones

Calculation

Combinations

7

All different

7!

5040

6

1 1 1 1 1 2

7!/2!

2520

5

1 1 1 2 2

7!/2!/2!

1260

4

1 2 2 2

7!/2!/2!/2!

630

3

3 2 2

7!/3!/2!/2!

210

2

4 3

7!/4!/3!

35

...read more.

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