• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # We are investigating the number of different arrangements of letters.

Extracts from this document...

Introduction

EMMA’s Dilemma

We are investigating the number of different arrangements of letters.

Firstly we arrange EMMA’s Name.

1)EAMM 7)MAEM

2)EMAM 8)MAME

3)MEMA 9)AMME

4)MEAM 10)AEMM

5)MMEA 11)AMEM

6)MMAE 12)EMMA

Secondlywe arrange lucy’s name.

1)Lucy 12)Cyul 22)Yulc

2)Luyc 13)Culy 23)Ycul

4)Lycu 14)Culy 24)Yluc

5)Lcuy 15)Cylu 25)Ucyl

6)Lcyu 16)Clyu

7)Ulcy 17)Cuyl

8)Ucly 18)Yluc

9)Uycl 19)Yucl

10)Ulyc 20)Yclu

11)Uylc 21)Ylcu

From these 2 investigation I worked out a method:

Step1: 1234---Do the last two number first then you get 1243.

1243---Do the last three numbers and try the possibility. 1423. 1432. 1342. 1324, because the number 2 has been the first number of last three numbers, so we don’t do it again.

Step2: we have list all arrangements of 1 go front, so we do 2 go front. 2134 and we do same thing to it, it will like this:

2134---2143, 2143---2431,2413,2314,2341

Step3: We have finished 2 go first, then let’s do 3 go ahead.

3124---3142, 3142---3241,3214,3412,3421

Step4: We have finished 3 go ahead, then try 4

4123---4132, 4132---4231,4213,4312,4321

We have list all arrangement of 1234, use this method we can arrange the number which has 5 figures or more.

We are trying to work out a formula which can calculate the number of arrangement when we look at a number.

Let’s list all the arrangment for 1234:

1234 4123

1243 4132

1324 --- 6 arrangment 4231 ---- 6 arrangement

1342 4213

1432 4321

1423 4312

2134 3124

2143 3143

2341 ---- 6 arrangements 3241 --- 6 arrangements

2314 3214

2413 3412

2431 3421

So if we time 6 by 4, we would get 24, and we can get the total arrangements of 24.

Let’s try 5 figures:

12345 13245

12354 13254

12435 --- 6 arrangements 13452 --- 6 arrangements

12453 13425

12534 13524

12543 13542

14235 15432

14253 15423

14352 --- 6 arrangements 15324 ---- 6 arrangements

14523 15342

14532 15243

14325 15234

Middle

5 1*3*4*5

Let’s work out the formular:

if n= number of figures

a= number of arrangements

the formular is a=ni/2

Let’s confirm the formular:

2 fig with 2 same number formular: 2/2=1 it works

3 (1*2*3)/2=3 it works

4 (1*2*3*4)/2=12 it works

Formular is confirmed

What about if 3 numbers are the same

let’s try 333

only on arrangement

Try 3331

3331

3313 ------ 4 arrangements

3133

1333

Try 33312

33312 31233 12333

33321 31323 13233 ---4 arrangements

33123 31332 ----12 arrangements 13323

33132 32331 13332

33231 32313

33213 32133

21333

23133----4 arrangements

23313

23331

Total arrangements are 4*5=20

Let’s try 6 fig with 3 same number

333124 332134

333142 332143

333214 334321

333241 332314

333412 332341 -----24 arrangements

333421 332413

331234 332431

331243 334123

331324 334132

331342 334213

331423 334231

331432 334312

312334 321334

312343 so on ----12 arrangements

312433

313234

313243 34-------

313324 --- 12 arrangements so on -----12 arrangements

313342

313423

313432

314233

314323

314332

123334 133324 2----

123343 133342 so on --------20 arrangements

123433 133234

124333 133243

124332 133423 --- 20 arrangements 4----

134323 133432 so on ---------20 arrangements

134233 142333

132334 143233

132343 143323

132433 143332

Total arrangement for 6 figure with 3 same number is 120, 20*6

Let’s see the construction:

3 fig with 3same number 1 arrangement

4 1*4

5 1*4*5

6 1*4*5*6

Can you see the pattern?

so the formula for three sames numbers of a number is:

a= ni/6

let’s review the formula:

formula for different number:

a=ni

formula for 2 same number:

a=ni/2

formular for 3 same number:

a=ni/6

Let’s put them is this way:

 n 1 2 3 x 1 2 6

n represent the number of figures of a number

x represent the divided number in the formular

Do you notice that x equal the last x time n, so I expect the formula for 4 sames number of a number is:

a= ni/24

Let’s confirm it:

try 4 same number.

4 fig, one arrangement.

a=n/24=(1*2*3*4)/24=1 the formular works

try 5 figures

11112

11121

11211 ----- 5 arrangements

12111

21111

a=n/24=(1*2*3*4*5)/24=5 the formular works

So formula is confirmed

Let’s investigate the formula, and improve it.

 n 1 2 3 4 5 x 1 2 6 24 110

Conclusion

Let’s confirmed

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.

Formular is confirmed

What about three pairs of different number of figures of a number

For example:

122333

according the formula, the total arrangment is

a=(1*2*3*4*5*6)/(1*1*2*1*2*3)=60

Let’s confirm it:

122333 212333 231332 3--------

123233 213233 232133 so on ------- 30 arrangements

123323 213323 232313

123332 213332 232331 -----30 arrangements

132233 221333 233123

132323 223133 233132

132332 223313 133213

133223 223331 233231

133232 231233 233312

133322 231323 233321

The formular works

Formular is confirmed

From the investigation above we find out the formular for calculating the number of arrangements, it’s

a=ni/xi

a represent the total arrangements

n represent the number of figures of the number

I represent the key I

x represent the numbers of figures of same number of the number

if there are more than one pair of same number, x2, or x3, so on may added to the formular, it depend how many pairs of same number.

For example:

for 2 pairs of same number of figures of same number of a number

the formula is a=ni/xixi

for 2 pairs of different number of figures of same number of a number

the formula is a=ni/x1ix2i

for 3 pairs of same number of figures of same number of a number

the formula is a=ni/xixixi

form 3 pairs of different number of figures of same number of a number

the formular is a=ni/x1ix2ix3i.

The formular can be also used to the arrangements of letter.

For example:

xxyy

the arrangement for this is a=(4*3*2*1)/(2*1*2*1)=6

xxyyy

the arrangement for this is a=(5*4*3*2*1)/(3*2*1*2*1)=10

xxxxxxyyyyyyyyyy

the arrangement for this is

a=ni/x1ix2i=(16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(10*9*8*7*6*5*4*3*3*2*1*6*5*4*3*2*1)=8008

The total arrangement is 8008.

Use this formular, we can find out the total arrangements of all numbers and 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

of Y's: Number of Y's Number of different combinations: 0 1 1 2 2 3 3 4 4 5 5 6 Rule: To find the number of all the different combinations possible, from one X, and a certain number of Y's ( using every letter only once, )

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

Let's try 5 figures: 12345 13245 12354 13254 12435 --- 6 arrangements 13452 --- 6 arrangements 12453 13425 12534 13524 12543 13542 14235 15432 14253 15423 14352 --- 6 arrangements 15324 ---- 6 arrangements 14523 15342 14532 15243 14325 15234 Don't you notice the arrangements of last 4 numbers added

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

BDCA CDBA DCBA With the four letter word above, I didn't need to retype this combination of letters ( four letters, all different ), but to make sure that it didn't matter which letters were used, I decided to try them again.

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

This is because if you had a word for example: EMMA there are 12 different arrangements, this is only because the Ms and repeated so for example if the on arrangement was EMMA and another EMMA although when you use two different colours it counts as two arrangements but when

1. ## I am investigating the number of different arrangements of letters in a word.

divided by 2 factorial (2) equals 12. Another example would be 6 letter word with 4 letters the same (6!/4!) which is equal to (720 divided by 24)=30. A general formula for this would be number of letters factorial divided by letter repeated factorial (l! /r!) I then investigated the number of arrangements with 2 or more letters the same.

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

Is simply 2 x 1 and the x 1 does not effect the answer. Therefore the formula for words with 3 letters the same is: n!/3! Prediction At this point it is becoming clearer to me the pattern between all the different numbers and formulae.

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

For example: Let's take the word 'end' for example. It has 3 letters in the word and so 3! should give us the total amount of different arrangements the word has. 3! = 3*2*1 3! = 6 We can prove this by writing out all the combinations for the word 'end.'

2. ## Investigate the number of different arrangements of the letters in a name.

JASME 94. EJASM 95. JMASE 96. EJMAS 97. SEJMA 98. ASEJM 99. MASEJ 100. MASJE 101. EMASJ 102. JEMAS 103. SJEMA 104. ASJEM 105. ASJME 106. EASJM 107. MEASJ Page 6 108. JMEAS 109. SJMEA 110. SJMAE 111. ESJMA 112. AESJM 113. MAESJ 114. JMAES 115. JAMSE 116. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 