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From Emma's name, a four letter word, we this time only get twelve arrangements. This exactly half of twenty-four,(twenty-four being the total number of arrangements for a four letter word with all the letters differ

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Introduction

Coursework

Part 1:

Lucy  Lyuc  Lcyu  Lcuy  Lycu  Luyc

Ulcy  Uylc  Ucyl  Uycl  Ulyc  Ucly

Cluy  Cylu  Cuyl  Cyul  Clyu  Culy

Yluc  Yclu  Yucl  Ycul  Ylcu  Yulc

If a word has: 2 letters it has 2 combinations. E.g. AB, BA

                     3 letters it has 6 combinations. E.g. ABC, ACB, BAC, BCA, CAB, CBA

The 6 combinations are found by multiplying the 3 letters times the 2 previous combinations. This works because the combination of the 2 letters only has 2 different combinations and because now there is another letter I just times the amount of letters in the new word by the amount of combinations in the previous word to find the amount of combinations in the new word.

So if the number of combinations to a 3 letter word I multiply together 3 and 2 too give the answer it becomes 3x2x1. And to get the amount of combinations of a for a 4 letter word I multiply together 4x3x2x1 to get the number of combinations. And that continues and work, but only when all the letters in the word are different.

...read more.

Middle

This formula is demonstrated in the table below.

Number of letters

Number of letters repeated

Arrangements

Calculation

2

2

1

2! divided by 2

3

2

3

3! divided by 2

4

2

12

4! divided by 2

5

2

60

5! divided by 2

6

2

360

6! divided by 2

7

2

2520

7! divided by 2

Part 3:

However, this formula was ineffective when trying to work out the number of arrangements for a word, which has three letters the same in it, and all the rest different.

AAAB   AABA

ABAA   BAAA

The word produced four arrangements, which is 24 divided 6(twenty-four being the number of arrangements of a four letter word with all the letters different, and six being the total number of arrangements of a three letter word and three is the number of letters repeated.)  I worked out that the correct formula for this was…

4!

6

because if all the A’s were different e.g. A(1) A(2)  A(3)

                                             A(1) A(3)  A(2)

                                                                 A(2) A(1)  A(3)

                                             A(2) A(3)  A(1)

                                             A(3) A(2)  A(1)

                                             A(3) A(1)  A(2)

These would all be different arrangements but seeing as they aren’t they have to be cancel out from 4!

...read more.

Conclusion

Arrangements:

XXYY

YYXX

XYXY

YXYX

XYYX

YXXY

Total Number of Arrangements: 6

Name: “XXXYYY”

Number of letters: 6

Arrangements:

XXXYYY

XYYXYX

YYYXXX

YXXYXY

XXYYYX

XYYXXY

YYXXXY

YXXYYX

XYYYXX

XYXXYY

YXXXYY

YXYYXX

XXYXYY

XYXYXY

YYXYXX

YXYXYX

XXYYXY

XYXYYX

YYXXYX

YXYXXY

Total Number of Arrangements: 20

Name: “XXYYY”

Number of letters: 5

Arrangements:

XXYYY

XYXYY

XYYYX

YYXXY

YYXYX

XYYYX

XYYXY

YYYXX

YXYYX

YXXYY

Total Number of Arrangements: 10

Summary of Part 4:

When there is only one letter repeated, the formula obtained was n!/y! (y being the number of repeated letters)

Therefore, for words where more than one letter is repeated, I will try the formula

image00.png

Examples:

Name: “XXXYYY”

Number of letters: 6

image01.png

Name: “XXYYY”

Number of letters: 5

image02.png

This equation appears to be correct.

  • We have therefore arrived at the formula image00.png
  • The n! is there as the total number of letters, as before. We previously decided that where one letter is repeated a certain number of times, the equation would be image03.png
  • It is therefore logical to assume that where more than one letter is repeated, n! should be divided by the total number of the first repeated letters multiplied by the total number of the second repeated letters.  So the over all formula is n!/X! x Y! e.c.t.

...read more.

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