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# Emma's Dilemma

Extracts from this document...

Introduction

## EMMA’S DILEMMA

1.

### No. of arrangements

Arrangements

1

EMMA

2

MEAM

3

AEMM

4

MMEA

5

MMAE

6

AMEM

7

EMAM

8

AEMM

9

MAEM

10

AMME

11

MAME

12

MEMA

2.

 No. of arrangements Arrangements 1 LUCY 2 LYCU 3 LYUC 4 LUYC 5 LCYU 6 LCUY 7 UCYL 8 UYCL 9 ULCY 10 ULYC 11 UYLC 12

Middle

YLCU

21

YUCL

22

YULC

23

YCLU

24

YCUL

3.

 No. of arrangements No. of letters Position(n) 1 A 1 2 AB 2 6 ABC 3 24 ABCD 4 120 ABCDE 5 720 ABCDEF 6 5040 ABCDEFG 7 40320 ABCDEFGH 8 362880 ABCDEFGHI 9 3628800 ABCDEFGHIJ 10

I have completed the table by using a method I found after finding a pattern in the first few results. For a four-letter word (all different letters) there are 24 arrangements as I found in Question 2 (LUCY). There are six combinations beginning with each letter, with this knowledge I think that a five-letter word (all different letters) would have 120 arrangements. To prove this answer/back it up I have found another pattern in the table.

4 letter word         1x2x3x4=24 (No. of arrangements)

Conclusion

3

2

Aabb

6

3

Aaabb

10

4

Aaaabb

15

5

Aaaaabb

21

6

Aaaaaabb

28

7

Aaaaaaabb

36

8

aaaaaaaabb

45

To further my investigation I have decided to investigate words with 3 different letters. First I will try to find the number of arrangements for the word abc using the formula then I will check it is correct by working it out myself, if the 2 results are the same then I will carry on finding the number of arrangements for words with 3 different letters in and produce a table.

3 letter word / 1a  x  1b  x  1c  = No. of arrangements

1x2x3        /1x1 x 1x1 x 1x1 = 6/1 = 6

1. abc
2. acb
3. bca
4. bac
5. cab
6. cba

Both methods produced the same number of arrangements (6).

 Position (n) Words arranged No. of arrangements 1 Abc 6 2 Aabc 12 3 Aaabc 20 4 Aaaabc 30 5 Aaaaabc 42 6 Aaaaaabc 56 7 Aaaaaaabc 72 8 aaaaaaaabc 90

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