• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
• Level: GCSE
• Subject: Maths
• Word count: 1192

# Investigate the number of different arrangements of letters in different words

Extracts from this document...

Introduction

Aim: To investigate the number of different arrangements of letters in a different words

First I am going to investigate how many different arrangements in the name LUCY, which has no letters the same.

LUCY

LUYC

LYCU

LYUC

LCYU

LCUY

ULCY

ULYC

UCLY

UCYL

UYLC

UYCL

CLYU

CLUY

CULY

CUYL

CYLU

CYUL

YLUC

YLCU

YULC

YUCL

YCLU

YCUL

There are 4 different letters and there are 24 different arrangements.

SAM

SMA

MSA

MAS

ASM

AMS

There are 3 different letters in this name and 6 different arrangements.

TO

OT

There are 2 different letters in this name and there are 2 different arrangements.

Obviously there is only 1 arrangement for a word with one letter.

 Number of Letters in a word Number of Arrangements 1 1 2 2 3 6 4 24

From the table of results I have found out that a 2 letter word has 2 arrangements, and a 3 letter word has 6.

Taking for example a 3 letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4! which is called 4 factorial.

Middle

6x5x4x3x2x1

720

7

7x6x5x4x3x2x1

5040

8

8x7x6x5x4x3x2x1

40320

9

9x8x7x6x5x4x3x2x1

362880

Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, 3 letters repeated and 4 letters repeated.

EMMA

AMME

AMEM

EMAM

AEMM

EAMM

MMEA

MMAE

MEMA

MAME

MEAM

MAEM

4-letter word, 2 letters repeated, 12 different arrangements.

MUM

MMU

UMM

3-letter word, 2 letters repeated, 3 different arrangements.

PQMMM

PMQMM

PMMQM

PMMMQ

QPMMM

QMPMM

QMMPM

QMMMP

MPQMM

MPMQM

MPMMQ

MQPMM

MQMPM

MQMMP

MMPQM

MMQMP

MMMPQ

MMMQP

MMPQM

MMMQP

5 letter word, 3 letters repeated, 20  different arrangements.

SMMM

MSMM

MMSM

MMMS

4 letter word, 3 letters repeated, 4 different arrangements.

RRRRK

RRRKR

RRKRR

RKRRR

KRRRR

5 letter word, 4 letters repeated, 5 different arrangements.

When looking at the number of arrangements for Emma I realized that there were 12 and this was half of the number of arrangements for Lucy so I decided to look at the others. This also was the case for the 3 letter words such as Mum and Sam.

24/2 = 12 – The number of arrangements for Emma.

6/2 = 3 – The number of arrangements for Mum.

I started to look at what they had in common apart from the divide by two.

Conclusion

As before, the original formula:

n! = the number of letters in the word

p! = the number of letters the same

From this I have come up with a new formula  The number of total letters factorial, divided by the number of x's, y's etc factorised and multiplied.

For the above example:

A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)

So : 1x2x3x4

(1x2) x (1x2)

= 24

4      = 6 different arrangements

A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y)

So: 1x2x3x4x5

(1x2x3x4) x (1)

= 120

24     = 5 different arrangements

A five letter words like abcde; this has 1 of each letter (no letters the same)

So : 1x2x3x4

(1x1x1x1x1x1)

= 24

1        = 24 different arrangements

A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's).

So : 1x2x3x4x5

(1x2x3) x (1x2)

= 120

12    = 10 different arrangements

This shows that my formula works:

n!   = the number of letters in the word

x!y! = the number of repeated letters the same

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. ## I am investigating the number of different arrangements of letters

Carry on, if a number has 7 fugure, then the total of arrangement should be 720 times by 7, and get 5040, the total of arrangement is 5040. This is my prediction, let's work it out a formula, and confirm it.

2. ## The relationships between the number of different spacers in an arrangement of square tiles ...

Of Triangles No. Of ^ Spacers No. Of K Spacers No. Of * Spacer 1 1 3 0 0 2 4 3 3 0 3 9 3 6 1 4 16 3 9 3 5 25 3 12 6 The rule for the ^ shaped spacer will always be three, because there are always three corners on a triangle.

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

This is due to the fact that in the name Lucy, all of the letters are different, so give a complete set of arrangements. Question Three: Choose some different names. Investigate the number of different arrangements of the letters of names you have chosen.

2. ## From Emma's name, a four letter word, we this time only get twelve arrangements. ...

combinations so the number of combinations for a five-letter word is divided by the amount of combinations the repeated letters make to get the answer. This can then be put into a general formula as........ n! y! ( 'y' equalling the number of repeated letters in the word)

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

For example: 1*2 = 2i 1*2*3 = 3i 1*2*3*4 = 4i so on So if n represent the number of figures of a number, then it has arrangements of ni. The formula: NI NI: Can be caculated on caculator. Process: pres key N (the number of figure), then press key I, then you would get the arrangements.

2. ## Investigating the arrangements of letters in words.

as this is the same as my previous formula. I will now test this and predict the combination for 5 different letters. 5 different letters 120 1x2x3x4x5=120 or 5! =120 I will now prove that 5 different letters has 120 combinations: ABCDE=24 ABCED=24 ABECD=24 AEBCD=24 EABCD=24 Total=120 I worked out the formula was n!

1. ## Investigate the number of different arrangements of letters in a word.

Like in Emma; it has 4 letters but 2 of which are the same. 4 factorial equals 24, but I could only find 12, which means that there are more to it than just factorial in that way. -To make it a bit easier instead of using letters I will use "x"'s and "y"'s (or any other letter any letter).

2. ## Investigate the number of different arrangements of the letters of Lucy's name.

To find out easier ways for finding different arrangements of various groups of letters I found out a relationship between the amount of letters and the number of arrangements. To do this I tested certain words ranging from 1 letter to 4 letters.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to