# Emma&#146;s Dilemma

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Introduction

Maths GCSE Coursework.

## Emma’s Dilemma

Emma is playing with the arrangements of the letters of her name. The standard arrangement is ‘EMMA,’ another arrangement is ‘MEAM,’ and a different arrangement is ‘EAMM.’

- Investigate the number of different arrangements of the letters of EMMA’s name.

Emma has a friend called LUCY

- Investigate the number of different arrangements of the letters of LUCY’s name.

Choose some different names

- Investigate the number of different arrangements of the letters of the names you have chosen.

A number of X’s and Y’s in a row such as : XXX……..XXYY……..YYY

- Investigate the number of different arrangements of the letters.

Shown below is the full list of the combinations of her name:

EMMA

EMAM

EAMM

AMME

AMEM

AEMM

MMAE

MAME

MAEM

MEMA

MEAM

MMEA

As you can see there are 12 combinations of Emma’s name. This process of writing all of the combinations out is very slow and very tedious; therefore it would be a lot better if we had a formula to work out the number of combinations. This would prove especially useful if you had a name like Bartholomew, which would take you ages to work out the number of combinations.

Middle

2

3

4

5

6

7

8

9

10

Combinations

1

2

6

24

120

720

5040

40320

362880

3628800

There is however the problem of repeating letters, like the two m’s in EMMA. This make the number of possible formulas a lot less, especially in longer names. However an example for 3 letters is shown below:

ABC

ACB

BAC

BCA

CAB

CBA

= 6 Combinations

ABB

BAB

BBA

= 3 Combinations

As you can see the number of combinations change, and we therefore need a different formula to work it out, or we need to make some modifications to our current one. I will begin trying to find this formula, by using just one repeating letter:

AB ABB ABBB ABBBB ABBBBB ABBBBBB

BA BAB BABB BABBB BABBBB BABBBBB

BBA BBAB BBABB BBABBB BBABBBB

BBBA BBBAB BBBABB BBBABBB

BBBBA BBBBAB BBBBABB

BBBBBA BBBBBAB

BBBBBBA

2 3 4 5 6 7

= No. Of Combinations

As you can see from above, it seems that the number of combinations is just the same as the number of letters; this however is only true for one repeating letter. When you get two repeating letters it changes quite considerably. If you take 4 letters, as in LUCY, there are 24 combinations ordinarily, with no repeating letters. If however there is one repeating letter as in EMMA there are only 12 combinations, if there are two letters which repeat, as in AABB, there would only be 6 combinations. As you can see from the table below the number of combinations halve each time, but this is not the case for all different names, or combinations of letters.

Number of Letters | 4 | 4 | 4 |

Number of repeated letters | 0 | 1, twice | 2, twice each |

Number of combinations | 24 | 12 | 6 |

Conclusion

= (120) / (2 * 6)

= 10

(This works for this one but I will now try it for my other results to make sure that it is right.)

For XXY:

Combinations = 3! / (2! * 1!)

= 6 / 2

= 3 (The correct answer)

For XXXYYY:

Combinations = 6! / (3! * 3!)

= 720 / 36

= 20 (Again the correct answer)

As you can see from the calculations above, the formula to work out the number of combinations for 2 repeating letters is:

Number of Combinations = Total! / (No. of the one letters! * No. of the second letters!)

Or

C = T! / ( R! * R’!)

Where C is the total number of combinations, T is the total number of letters, R is the first repeated letters and R’ is the second repeated letters.

I have also discovered that by just adding on more R’’, and R’’’ you can work out the combinations for words or names with many repeating letters or just for all names by using the one formula. For example I know that AABBCC has 90 combinations and I will now prove it using the formula.

Combinations = T! / (R! * R’! * R’’!)

= 6! / (2! * 2! * 2!)

= 720 / 8

= 90 combinations.

I will now show how the formula can be applied to any name by incorporating all the letters, for STEVEN:

Combinations = T! / (No. of S’s! * No. of T’s! * No. of E’s! * No. of V’s! * No. of N’s!)

= 6! / (1! * 1! * 2! * 1! * 1!)

= 720 / 2

= 360 combinations of my name.

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

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