# Emma's Dilemma

Extracts from this document...

Introduction

27th June 2001 Tom Pountain 10A | |||||||||||

Emma's Dilemma | |||||||||||

I investigated the number of different arrangements of four letters with no repetitions. 1)ABCD 2)ABDC 3)ADBC 4)ADCB 5)ACBD 6)ACDB 7)BACD 8)BCAD 9)BDCA 10)BADC 11)BDAC 12)CDBA 13)CBAD 15)CDAB 16)CADB 17)CBDA 18)DABC 19)DBCA 20)DCAB 21)DACB 22)DBAC 23)DCBA 24)DABC 25)BCDA | |||||||||||

I have found 24 different arrangements of these letters and this result is confirmed in the tree diagram. Secondly, I have investigated the number of different arrangements of four letters with one letter repeated twice. 1)ACBB 2)ABCB 3)BABC 4)BACB 5)BBAC 6)BBCA 7)BCAB 8)BCBA 9)CBBA 10)CABB 11)CBAB 12)ABBC I have found 12 different arrangements of the letters and this result is confirmed in the tree diagram. From these two investigations, I have worked out a method that can be used for further work: Firstly, with ABCD you rotate the last two letters, then you get ABDC. Then, with ABCD you must then rotate the last three letters and try the possibilities of ADBC, ADCB, ACDB, ACBD. Because the letter 'B' has been the first number of last three letters before, we don’t do it again. We have list all arrangements with A first, so now we do B secondly: BACD, and we do same thing to it, it will like this: BACD=BADC, BADC=BDCA...BDAC...BCAD...BCDA We have finished with B first, then we do C next: CABD=CADB, CADB=CBDA...CBAD...CDAB...CDBA We have finished with C first, then we do D DABC=DACB, DACB=DBCA...DBAC...DCAB...DCBA |

Middle

I have investigated the arrangements of ABBC:

1.ABBC

2.ABCB

3.ACBB

4.BBCA

5.BCAB

6.BCBA

7.BABC

8.BACB

9.BBAC

10.CABB

11.CBAB

12.CBBA

I have found 12 different arrangements of these letters. From the last investigation, I have decided to look at the last results, in this case '3' for a pattern. In relation to '3', 12 = 3x4, 4 being the number of letters in the arrangement.

I have investigated the arrangements of ABBCD:

1 12234

2 12243

3 12324

4 12342

5 12423

6 12432

7 13224

8 13242

9 13422

10 14223

11 14232

12 14322

12x5 = 60

I have found 60 (12x5) different arrangements of these letters.

Now I can show alternative data, after learning from the last investigation, for an arrangement with one repetition of 2 of the same letters.

3 = 1x3

4 = 1x3x4

5 = 1x3x4x5

They are not multiplied by 2 because 2 of the numbers (those repeated) can only be used once each in a certain space.

I can now try and find a formula in terms of n!.

if n = number of figures, and a= number of arrangements

the formula is: a = n!/2

This is because the lack of 'x2' in the equation acts the same as '/2' so the formula would be just n! With that adjustment. This can be seen when looking at the tree diagrams.

The formula can be confirmed when I look at the following data.

2 (1x2)/2 = 1

3 (1x2x3)/2 = 3

4 (1x2x3x4)/2 = 12

The reason for this formula is that it would be n!, for the same reason as aforementioned with:

But now, because of the two B's, there would be some arrangements from the tree diagram that are the same. I.e., ABBA and ABBA.

Conclusion

For example:

122333

according the formula, the total arrangement 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 formula works

Formula is confirmed

From the investigation above we find out the formula 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 formula, 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 formula is a=ni/x1ix2ix3i.

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

For example:

xxyy

the arrangement for this is a=(4x3x2x1)/(2x1x2x1)=6

xxyyy

the arrangement for this is a=(5x4x3x2x1)/(3x2x1)x(2x1)=10

xxxxxxyyyyyyyyyy

the arrangement for this is

a=n!/x1!x2!=(16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1)/(10x9x8x7x6x5x4x3x2x1)x(6x5x4x3x2x1)=8008

The total arrangement is 8008.

Use this formula, 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.

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