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

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Introduction

Emma's Dilemma

Question One:

Investigate the number of different arrangements of the letters of Emma's name.

Question Two:

Emma has a friend named Lucy, Investigate the number of different arrangements of the letters of Lucy's name.

Question Three:

Choose some different names. Investigate the number of different arrangements of the letters of names you have chosen.

Question Four:

A number of X's and a number of Y's are written in a row such as

XX………XXYY………Y

Investigate the number of different arrangements of the letters.

Question One:

Investigate the number of different arrangements of the letters of EMMA's name.

Answer:

In order for me to answer this question, I will write down all of the different arrangements for the letters of Emma's name. This will allow me to get the total number of arrangements, which will help me to find a rule in the latter questions in this piece of coursework.

Below are all of the different arrangements of the letters of Emma's name:

EMMA

EMAM

EAMM

MEMA Total:

MEAM 12

MAEM

MAME

MMEA

MMAE

AMME

AMEM

AEMM

From my results, I can see that there are 12 different combinations of the letters from Emma's name, if all of the letters are used, and only once.

I have set out the letters in an ordered fashion, so that it is easier to find all of the combinations, and to only do a combination once.

Therefore, my findings for a four letter word ( EMMA ), if all letters are used and only once, are that there are only 12 different arrangements. This is due to the fact that in the name "Emma", there are two "m's", which reduce the number of arrangements.

This id due to the

Middle

Results:

Number of Letters: Number of different combinations:

4 6

5 30

Rule:

To find the number of all the different combinations possible, from a selected number of letters ( using every letter only once, ) with two letters being chosen twice, can be achieved by using the following formula:

To find the total number of combinations, take the number of letters, and times that number by the number of letters minus 1, timed by the number of letters minus 2, and so on, until the number being times by is the number 1 it-self, and then divide it by the 2 ( the number of times 'A' has been repeated ) times 2 ( the number of times B has been repeated ).

No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc…

combinations ( times a letter X ( times a different letter

has been repeated ) has been repeated )

For example:

To find the number of combinations which can be made from using 10 letters, with two of the ten having two of the same letters, can be found by using the following formula:

No. of = 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1

Combinations 2 X 2

= 907200

Justification

The reason why this rule occurs, is because there may-be the same number of letters, but two are used twice. This means that the total number of arrangements possible is reduced by a factor of 4, or "2 ! X 2!" ( See later ).

For example

For 6 letters, all different, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.

For 6 letters, 2 repeated twice, there are "6 X 5 X 4 X 3 X 2 X 1"arrangements.

2 X 2

This is due to the fact that the six different letters can be arranged in any order, and because all of them are different, it dose matter which order they go in.

Conclusion

Number of Y's: Number of different combinations:

0 1

1 2

2 3

3 4

4 5

5 6

Table 7, Two X's + increasing number of Y's:

Number of Y's: Number of different combinations:

0 1

1 3

2 6

3 10

4 15

Table 8, Three X's + increasing number of Y's:

Number of Y's: Number of different combinations:

0 1

1 4

2 10

3 20

4 35

Conclusion

From my in-depth study into factorials, X's and Y's, and many more things, I conclude that the easiest, and simplest rule to find the number of arrangements for any set of letters, is the one I showed, and explained at the beginning of this piece of coursework.

This rule is:

D = T ! ( n + p ) !

n ! X p ! or n ! X p !

where,

D = number of different arrangements,

T = total number of letters,

n = number of letter X's, and

p = number of letter Y's.

I am now confident that I can extend this rule to even larger combinations of letters, as follows:

D = T !

m ! X n ! X p ! … X z !

Where lower case letters m, n, p, … z are the different numbers of each different letter used, and

T = m + n + p + … + z

Table at the end of this piece of coursework

Finally, I have also attached a tables, to make finding lower number of X's and Y's easier, and quicker to find. To use it, just count the total number of letter , and then the number of times the letter Y, and letter X have been repeated.

For example, to find the number of arrangements for:

XYYXYXXXYY

You count up the total number of letters, which is 10, then the number of Y's, which is 5, and finally the number of X's, which is also 5.

Letters = 10

Y's = 5

X's = 5

With these numbers, you need to find the right axis's. Then, read along the "X" axis, until you get to the number 5 ( the number of X letters for this set of letters ). Finally, read up the "Y" axis, until you get to the number 5 ( the number of Y letters for this set of letters ). Where the two numbers cross, is the total number of arrangements for that set of 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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