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• Level: GCSE
• Subject: Maths
• Word count: 2883

# GCSE Mathematics: Emma's Dilemma

Extracts from this document...

Introduction

GCSE Mathematics: Emma's Dilemma Emma and Lucy are playing with arrangements of their names. One arrangement of LUCY is L.U.C.Y. Another arrangement is Y.L.C.U. We are investigating the number of different arrangements of Emma and Lucy's names. Then we shall move onto the number of different arrangements of various groups of letters. Firstly, I shall list all the possible combinations for the word "LUCY": LUCY LCUY LYUC LYCU LUYC LCYU UCYL UYCL ULCY ULYC UYLC UCLY CYLU CYUL CULY CUYL CLYU CLUY YCUL YCLU YUCL YULC YLUC YLCU = 24 Combinations I listed all these combinations by using the following method: Step 1 Arrange the listing process into 4 stages: 1 _ _ _ 2 _ _ _ 3 _ _ _ 4 _ _ _ Step 2 You start off with the original word: 1234 Acquire the combinations of the last two numbers first and you end up with 1243. Now that you have 1243, do the last three numbers and try the different possibilities: 1423 1432 1342 1324. Because the number 2 has been the first number of last three numbers, we don't do it again. Step 3 Now that we have listed all the arrangements beginning with 1, we do the list of arrangements with 2 in the beginning: Start off with 2134 and do same thing to it, it will look like this: 2134 2143 2431 2413 2314 2341 Step 4 Now that we have established the different arrangements of 2, we go ahead and do 3 at the beginning: 3124 3142 3241 3214 3412 3421 Step 5 Now that we have established the different arrangements of 3, we go ahead and do 4 at ...read more.

Middle

Try 5 letters: 42213 12234 42231 12243 42123 12324 42132 12342 42321 12423 42312 12432 41223 13224 41232 13242 41322 13422 43122 14223 43212 14232 43221 14322 21234 23124 21243 23142 21324 23214 21342 23241 21423 23421 21421 23412 22134 24123 22143 24132 32412 22314 24231 32421 22341 24213 34122 22413 24312 34212 22431 24321 34221 32241 32214 32142 32124 31422 31242 31224 So the total arrangements are 12 x 5=60 Let's put our results in a table and try to extract a formula from there: Number Of Letters (With 2 of the letters the same) Number Of Combinations 2 1 3 6 4 12 5 60 Which can be re-written as: Number Of Letters (With 2 letters the same) Number Of Combinations 2 1 3 3 x 1 4 4 x 3 x 1 5 5 x 4 x 3 x 1 Let's work out the formula: if n= number of figures a= number of arrangements the formula is a=n!/2 Let's confirm the formula: Number Of Letters (with 2 letters the same) Number Of Combinations 2 2/2=1 It works 3 (1 x 2 x 3) / 2=3 It works 4 (1 x 2 x 3 x 4) / 2=12 It works Now I shall investigate different amounts of letters with different amounts of repeated letters. I shall start off with a four letter word where 3 of the letters are the same. For this I shall use the combination "3331": 3331 3313 3133 1333 - There are 4 combinations. Now let's try a 5 letter word with 3 letters the same: 33312 31233 33321 31323 33123 31332 33132 32331 33231 32313 33213 32133 21333 12333 23133 13233 23313 13323 ...read more.

Conclusion

The formula is confirmed. Conclusion From the investigation above we find out the formula for calculating the number of arrangements, it's a=n!/x! a represent the total arrangements n represent the number of figures of the number ! represents the factorial function. x represent the number of duplicated letters in the word for 2 pairs of the same letter: the formula is a=n!/x!x! for 2 pairs of different duplicated letters of same number in a word: the formula is a=n!/x1!x2! for 3 pairs of same number of duplicated letters in a word: the formula is a=n!/x!x!x! for 3 pairs of a different number of the same letter in a word: The formula is a=n!/x1!x2!x3!. The formula can be also used to find the amount of different arrangements of letters. For example: xxyy the arrangement for this is a=(4 x 3 x 2 x 1)/(2 x 1 x 2 x 1)=6 xxyyy the arrangement for this is a=(5 x 4 x 3 x 2 x 1)/(3 x 2 x 1 x 2 x 1)=10 xxxxxxyyyyyyyyyy The formula for this is: a=n!/x1!x2!=(16 x 15 x 14 x 13 x 12 x 11 x 10 x 9x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ) / ( 1 0 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 3 x 2 x 1 x 6 x 5 x 4 x 3 x 2 x 1 ) = 8 0 0 8 The total arrangement is 8008. Using this formula, we can find out the total arrangements of all numbers and letters. Sheroy Zaq (X2) Mrs Whybrow ...read more.

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1. ## Maths GCSE Coursework: Emma's Dilemma

We can see here that when another pair of the same letter comes into the case, the result is exactly half than what it was before the pair was added (shown in bold RED). Why this works Explaining why this works is difficult, so I will use the help of the previous studies to explain.

2. ## Emma's Dilemma

If you wanted to know the number of different arrangements for a 5-letter word with one letter repeated: * 2? - you would calculate 5! ? 2. * 3? - you would calculate 5! ? 6. * 4? - you would calculate 5!

1. ## EMMA's Dilemma Emma and Lucy

112233 121233 123123 131223 132231 112323 121323 123132 131232 132123 112332 121332 123213 131322 132132 ------ 30 arrangements 113223 122133 123231 132321 133122 113232 122313 123312 132312 133212 113322 122331 123321 132213 133221 2------- 3------ so on ------30 arrangements so on --------- 30arrangements The total arrangement is 90, the formular works.

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

so on, until the number being times by is the number 1 itself. Then divide the number by the number of times the letter has been repeated. No. of = No. of letters X ( No. of letters - 1 )

1. ## I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

I worked out that the total number of arrangements was worked out by multiplying the number of letters by all the numbers under that number. So if I was to work out the number of arrangements for a 4 letter word, I would multiply 4 by 3 by 2 by 1, as they are the numbers below that number.

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RREYK KYRER ERKYR RYKRE RRYKE KYERR ERKRY RYKER RRYEK KRERY EYRRK RERKY RKERY KRREY EYRKR RERYK RKEYR KRRYE EYKRR REYKR RKYRE KREYR EKRRY REYRK RKYER KRYRE EKRYR REKRY RKREY KRYER EKYRR REKYR RKRYE YKERR YERRK YRREK YREKR YKRER YERKR YRRKE YRKRE YKRRE YEKRR YRKER YRERK There are 60 combinations for a five-letter name with one repeated letter.

1. ## We are investigating the number of different arrangements of letters.

According to the formula, I expect the total arrangement for this is a=(1*2*3*4*5)/(3*2*1*2*1)=10 11122 12211 11212 21112 11221 21121 -------10 arrangements 12112 21211 12121 22111 The formular still works. Let's try 7 fig, with 3 same number, and 4 same number.

2. ## Emma's dilemma

ADBCE 15) ADCBE 16) ADCEB 17) ADECB 18) ADEBC 19) AEBCD 20) AEBDC 21) AEDBC 22) AEDCB 23) AECDB 24) AECBD There are 24 arrangements starting just with one letter "A". therefore there are five letters altogether, so 5*24=120 Consequently there are 120 arrangements altogether. 51 = 5*4*3*2*1 = 120 arrangements.

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