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• Level: GCSE
• Subject: Maths
• Word count: 1007

# Emma's Dilema

Extracts from this document...

Introduction

Maths coursework

Emma and Lucy are playing with arrangements of their names, they want to know how many different arrangements each of their names has; and how to work out other words, number of arrangements.

LUCY arrangementsEMMA arrangements

LUCY                                                EMMA

LUYC                                                EMAM

LCYU                                                EAMM

LCUY

LYUC                                                MAME

LYCU                                                MAEM

MEAM

ULCY                                                MEMA

ULYC                                                MMAE

UCYL                                                MMEA

UCLY

UYCL                                                AMME

UYLC                                                AMEM

AEMM

CULY

CUYL

CLYU

CLUY

CYLU

CYUL

YCUL

 No. of letters No. of repeats No. of arrangements LUCY 4 0 24 EMMA 4 1 12

YCLU

YUCL

YULC

YLUC

YLCU

LUCY has 24 different arrangements for her name; whereas EMMA, with one repeat, only has 12. The equation (number of letters ×6)÷(number of repeated letters) works for these two words.

Middle

0

yes

EMMA

4

1

yes

NOON

4

2

no

Rather than trying to guess random formulas and try to make them work I will, get more data.

TIM                                                                        BOB

TMI                                                                        BBO

IMT                                                                        OBB

ITM

MTI

MIT

(Number of letters×2)÷(Number of repeated letters) works for the 3 letter words but can’t work for the 4 letter words because 6× does.

 No. of letters No. of repeats Multiplier of letter no. LUCY 4 0 6 EMMA 4 1 6 NOON 4 2 ? TIM 3 0 2 BOB 3 1 2

The formulae can’t work for any other letter words, and the 4 letter word equation didn’t work for 2 repeats in a 4 letter word. So forget repeats first, I’ll see if I can find an equation for no repeats that works for all by focussing on no repeated lettered words.

 No. of letters No. of arrangements TIM 3 6 LUCY 4 24

LUCY has 4 letters if I multiply that by the number of different arrangements of the 3 letter word TIM I get the number of different arrangements for LUCY.

Conclusion

24÷6. I had to use factorials before, and it seems like they are needed again. As said earlier 6 can be expressed as 3!, so number of letters!÷ number of repeats!.

A five lettered word with one repeated letter has 60 arrangements. So 5!(120)÷2!(2)=60.So this works, what about 5 lettered arrangements with 2 repeats.

A(BBBC)4

B(ABBC)12

B(ABBC)12

B(ABBC)12

C(ABBB)4_

44

Two of them are the same so ignore them, so a 5 lettered with 2 repeats would have 20 arrangements. 5!÷3!=20. So this also works. I will do one more test to make 100% sure it works. 6 lettered words have 720 arrangements (6×120). So:

A(BBCDE)60

B(ABCDE)120

B(ABCDE)120

C(ABBDE)60

D(ABBCE)60

E(ABBCD)60_

480

Ignore one of the 120, so a 6 lettered word with one repeat has 360 arrangements. 6!(720)÷2!(2)=360. This is also correct!

So to find the number of arrangements of any word you need to use the equation. n!÷r!Where n is the number of letters in the word and r is the amount of times a number is repeated. I.e. In EMMA n=4 and r=2!.

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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