# Emma&#146;s Dilemma

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Introduction

Thomas Collyer 10Lr MATHS

27/04/07 EMMA’S DILEMMA

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## EMMA’S DILEMMA

## Aim

To investigate the patterns caused by the permutations of letters in words of different lengths and to investigate the possibility of predicting the number of permutations. To discover a formula that can be applied to all words.

Question 1.

12 different permutations can be made by the name EMMA:

- EMMA
- EMAM
- EAMM
- MMEA
- MMAE
- MEMA
- MEAM
- MAEM
- MAME
- AMEM
- AMME
- AEMM

Question 2.

Middle

- ELICV
- ELIVC
- ELCVI
- ELCIV
- EICVL
- EICLV
- EILVC
- EILCV
- EIVCL
- EIVLC
- EVCIL
- EVCIL
- EVLIC
- EVLCI
- EVILC

120.EVICL

4) Looking at mathematics behind answers

The word Lucy has 4 letters l,u,c and y. If we know how many letters (no repeats) are in a word we can work out the number or permutations by multiplying 1 x 2 x 3 x 4 and so on until we reach the correct number of digits in a word (e.g. a 10 letter word would be 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10). The mathematical symbol for this is n! and stands for number factorial, this shows number of letters! . After working this out we can quickly and easily work out the number of permutations made by a word with repeats – this id done by dividing our previous answer by 2. For 2 repeats we divide the answer again by 2.

Conclusion

(In words: n factorial over p factorial times q factorial and so on)

Example1

How many permutations can be formed from the letters, taken 5 at a time, of the word DADDY?

Solution (has 3 letters that are the same – D)

5! / 3! = 20

- The 5! Shows the word has five letters in it including repeats.
- The 3! Shows the word has 3 letters that are the same (d)

Example2

Find the number of arrangements of all the letters in the word MARMALADE.

Solution

9! / 2! x 3! = 30240

- The 9! Shows the word has 9 letters altogether including repeats.
- The 2! Shows the word had 2 letters the same (m).
- The 3! Shows the word has 3 more letters that are the same (a).

This formula can be used conclusively for other problems.

In how many ways can 3 apples, 2 oranges, 4 pears and one banana be given to 10 children if each child receives a piece of fruit?

Solution

10! / 3! x 2! x 4! = 12600

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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