• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Emma's Dilemma

Extracts from this document...

Introduction

Mathematics GCSE

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

...read more.

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)

...read more.

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

...read more.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Emma's Dilemma

    following it: 3 letters: A/A/B AAB ABA BAA 3 combinations 4 Letters: A/A/B/C AABC AACB ABCA ABAC ACBA ACAB BAAC BACA BCCA CAAB CABA CBAA 12 combinations (2 pairs of 2 different letters) I will now look at two pairs of 2 different letters with nothing following it: 4 letters:

  2. GCSE Mathematics: Emma's Dilemma

    Formula for a word with 2 of the same letter: a=n!/2 Formula for a word with 3 of the same letter: a=n!/6 I shall put them in a table for easier analysis: n 1 2 3 x 1 2 6 n represent the number of figures of a number x represent the number which n!

  1. Emma's Dilemma

    Now I will investigate another 6-letter word with 2 different letters, this time with 1 repeated 4 times, 1 repeated twice. XXXXYY XXYYXX YXXXXY XXXYXY XYXXXY YXXXYX XXXYYX XYXXYX YXXYXX XXYXXY XYXYXX YXYXXX XXYXYX XYYXXX YYXXXX There are 15 different arrangements.

  2. Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

    Algebraic Rule Due to this new mathematical symbol I have decided to use, I can now right the previous equation even more simply. If we take: "n" to be the number of letters, "A" to be any repeated letter, "B" to be any repeated letter, and "C" to be any

  1. Emma’s Dilemma.

    2 1 3 4 U L C Y 8 2 1 4 3 U L Y C 9 2 3 1 4 U C L Y 10 2 3 4 1 U C Y L 11 2 4 1 3 U Y L C 12 2 4 3 1 U

  2. Emma’s Dilemma

    I applied this to my rule. L x no of arrangements for one letter = A 5 x 24 = 120 This gives me all the information that I need about one, two, three, four and five letter names with all different letters.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work