• 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

Danny Soopen –“Emma’s Dilemma” - Page  of

Danny SoopenGCSE Maths Project – “Emma’s Dilemma”

I have been told to investigate the number of different arrangements of the letters of 'EMMA'. I was then told to investigate the number of different arrangements of the letters of 'LUCY'. After these, I must investigate into different arrangements of letters and find a rule to find the number of different arrangements of letters of any name.

Part 1:

Ways of arranging EMMA’s name:

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

For each new beginning letter, there are 3 different possible combinations.

There are a total of 12 possible different combinations.

Part 2:

Ways of arranging LUCY’s name:

 LUCY ULCY CULY YUCL LUYC ULYC CUYL YULC LCUY UCLY CLUY YCUL LCYU UCYL CLYU YCLU LYUC UYLC CYLU YLUC LYCU UYCL CYUL YLCU

For each new beginning letter, there are 6 different possible combinations.

There are a total of 24 possible different combinations.

This combination consists of all the letters being different. I will also try and find a formula for this arrangement, and number of letters. Once I have discovered these formulae I am going to investigate, other combinations of letters and different amounts of letters. I will then try to discover a link between the formulae to enable me to find a formula for the general case. There are 24 different possibilities in this arrangement of 4 letters all different. I have noticed that with Lucy beginning each different letter. For example there are 6 arrangements with LUCY beginning with L, and 6 beginning with u and so on 6*4(the amount of letters) gives 24.

Middle 1          1          1          1

We can say that  tn = n!

For example:

For FRANK’s name,         tn = 5!

tn = 5 x 4 x 3 x 2 x 1

= 120

The formula for the number of combinations for a word where all the letters are different can be obtained, (as with the LUCY example on page 2) by simply writing out all of the permutations. As no letters are repeated, it is a straightforward formula.

There are a total of four letters to be placed into four different positions each time. When the first letter is placed, there are four different places to put it in. When the second letter is being placed, one position is already being taken up by the first letter, and so there are only three places where it can be positioned. When the third letter is being placed, there are only two places in which it can be placed, and the final letter must fill the final space.

In the formula, we start with the number four as there are four letters in the word, and at each stage of the arrangement process, there is one less available place in which to put the letters, so the multiplying number decreases at each stage

Therefore, where there are four letters to be placed, we can summarise it as:

tn = 4 x 3 x 2 x 1

or        tn = 4!

(b).        Investigating the number of different arrangements of letters in names with varying numbers of letters, where one letter is repeated once.

Conclusion  where n = the number of letters in the word,

and R = the number of times the letter occurs.

Part 4:

Investigating the number of different arrangements of letters where more than one letter occurs more than once.

For example: words such as XXYY, XXYYXX, and XXXYYY

The above examples all have more than one letter that repeats more than one time.

Investigation:

Name: “XXYY”

Number of letters: 4

Arrangements:

 XXYY YYXX XYXY YXYX XYYX YXXY

Total Number of Arrangements: 6

Name: “XXXYYY”

Number of letters: 6

Arrangements:

 XXXYYY XYYXYX YYYXXX YXXYXY XXYYYX XYYXXY YYXXXY YXXYYX XYYYXX XYXXYY YXXXYY YXYYXX XXYXYY XYXYXY YYXYXX YXYXYX XXYYXY XYXYYX YYXXYX YXYXXY

Total Number of Arrangements: 20

Name: “XXYYY”

Number of letters: 5

Arrangements:

 XXYYY XYXYY XYYYX YYXXY YYXYX XYYYX XYYXY YYYXX YXYYX YXXYY

Total Number of Arrangements: 10

Summary of Part 4:

When there is only one letter repeated, the formula obtained was Therefore, for words where more than one letter is repeated, I will try the formula Where n = the total number of letters

And   R1 = the number of times the first letter occurs

And   R2 = the number of times the second letter occurs

Examples:

Name: “XXXYYY”

Number of letters: 6 This formula does not work. I think that the formula could perhaps be Examples:

Name: “XXXYYY”

Number of letters: 6 Name: “XXYYY”

Number of letters: 5 This equation appears to be correct.

• We have therefore arrived at the formula • The n! Is there as the total number of letters, as before. We previously decided that where one letter is repeated a certain number of times, the equation would be • It is therefore logical to assume that where more than one letter is repeated, n! Should be divided by the total number of the first repeated letters multiplied by the total number of the second repeated letters.

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

# Related GCSE Emma's Dilemma essays

1. ## Emma's Dilemma

1 letter: A A: 1 combination (1x1) 2 letters: A/B AB,BA: 2 combinations (2x1) 3 letters: A/B/C A (BC) B (AC) C (AB) Here we have three starting letters (A,B and C). We can do this systematically. We start with A. Now, because we have written A, we know that there can only be two more letters which go next, B and C.

2. ## Emma&amp;amp;#146;s Dilemma

Therefore 3 x 12 = 36 arrangements. M M E A G* So with 'M' as the starting letter there are 24 arrangements. Therefore 36 + 24 = 60 arrangements. Another Formula for this would be N! / 2 = a O R N! / X! = a Where N = the number of letters in the

1. ## Emma&amp;amp;#146;s Dilemma

/ r! = 5! / 3! = 120 / 6 = 20 this is also the right amount. This also works for when there are two repetitions: 4th = x! / r! = 4! / 2! = 24 / 2 = 12 and for when there is only one repetition of each letter (i.e.

2. ## Emma's Dilemma

This means that to get the number of different combinations for a word with 3 of the same letter in it you must have to dived the total number of combinations by 6 so that you eliminate those words where the letters are in the same order.

1. ## Emma's Dilemma

Now that I have worked out the how many possibilities there are for a 2 lettered word, a 3 lettered word, and a 4 lettered word, I think I may be able to figure a pattern out. Therefore I will list all my discoveries in a table on the following page.

2. ## Emma's Dilemma

Table of Results. Number of arrangements with Number of letters Double letters Single letters 1 1 2 1 2 3 3 6 4 12 24 I cannot see the formula yet, but I have noticed some patterns in the results.

1. ## EMMA's Dilemma Emma and Lucy

21243 23142 31242 21324 23214 31422 21342 23241 32124 21423 23421 32142 21421 23412 -----24 arrangments 32214 ------- 12 arrangements 22134 24123 32241 22143 24132 32412 22314 24231 32421 22341 24213 34122 22413 24312 34212 22431 24321 34221 so the total arrangements are 12*5=60 We have found the frequency 2

2. ## Emma's Dilemma

I considered two-letter names such as the "Al" and "Mo". I found that there were only 2 different arrangements for the letters in these cases: AL LA MO OM There was now definitely a correlation between the number of letters in a name and the number of ways in which • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work 