# How does Nike benefit from competitive advantage?

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Introduction

## Introduction

## In my investigation I am going to look at the number of different arrangements of letters in a word and the formulae that connect them. I am going to start by finding the formula for the number of arrangements for words which have all different letters. I am then going to continue by finding the formula for the number of arrangements for words with two letters the same. This will be carried on to find the number for different arrangements of various groups of letters and a formula will be determined that connects them all.

## Part 1

## Aim

## To look at words with no duplicated letters, finally investigating the number of different arrangements of the letters of Lucy’s name.

## With 1 letter in the word there is 1 arrangement.

## 1) K

This is because there is only 1 space for the letter to go.

1

With 2 letters in the word there are 2 different arrangements.

1) JO 2) OJ

This is because there are 2 choices for the first letter and 1 for the last.

2 1

With 3 letters in the word there are 6 different arrangements.

1) LIZ 2) LZI 3) ILZ 4) IZL 5) ZLI 6) ZIL

This is because there are 3 choices for the first letter, 2 for the next and then 1

3 2 1

No. of letters | Working | Arrangement |

1 | 1 | 1 |

2 | 2x1 | 2 |

3 | 3x2x1 | 6 |

From this initial assessment the rule for any size word combination, where all the letters are different, is a= n! where n is the number of letters in the word and a is the total number of arrangements.

I would predict, therefore, that there will be 4! or 24 arrangements in the word Lucy.

Middle

Words with different letters | Words with 2 letters the same | Words with 3 letters the same | |

No. of letters | No. Arrangements | No. Arrangements | No. Arrangements |

3 | 6 | 3 | 1 |

4 | 24 | 12 | 4 |

5 | 120 | 60 | 20 |

6 | 720 | 360 | 120 |

With three letters the same (AAA) there is only one arrangement that is seen. However if they were labelled A1A2A3 we would see that there are in fact six (or 3!) combinations in practise. Consequently the equation becomes

a = n! / 3!

Where n = the number of letters in the word and a = the number of different arrangements.

To confirm the prediction I will look at a six letter word with one letter repeated three times. If the formula is correct there will be 6! / 3! different arrangements (i.e.120)

Assume the word is ABCDDD

If A B or C are the first letter there will be 3 x 20 arrangements as the last four letters will include one letter, D, three times. This is like AONNN where I have already shown there are 20 arrangements.

If D is the first letter there will be 5! / 2! arrangements (i.e. 60) as the last five letters include one, D, that is there twice ( see table above)

This gives a total of 120 different arrangements as predicted.

I therefore conclude that the connecting formula is

a= N! / r!

Where n = the number of letters in the word, r is the number of letters repeated and a = the number of different arrangements.

No. of letters | Words with different letters | Words with 2 letters the same | Words with 3 letters the same | Words with 4 letters the same |

No. Arrangements | No. Arrangements | No. Arrangements | No. Arrangements | |

4 | 24 | 12 | 4 | 1 |

5 | 120 | 60 | 20 | 5 |

6 | 720 | 360 | 120 | 30 |

Conclusion

Final Conclusion

From the work I have done I have developed a number of formula, some of which are for specific circumstances, leading to the universal formula which covers all possibilities.

- For words containing all different letters I have

a = n!

Where n = the number of letters in the word and a = the number of different arrangements.

This can also be written a = n! / 1! x 1! x 1!……….

- For words containing a letter repeated twice the equation is

a =n! / 2!

Where n = the number of letters in the word and a = the number of different arrangements.

This can also be written a = n! / 2! x 1! x 1! x 1!……

- For words containing a letter repeated three times the equation is

a = n! / 3!

Where n = the number of letters in the word, and a = the number of different arrangements.

This can be written as a = n! / 3! x 1! x 1! x 1!…..

The universal formula, covering all possibilities, is

a = n! / x! x y! x z!…….

Where n = the number of letters in the word, x y z…. are the number of similar individual letters and a is the number of arrangements.

n will be the sum of x + y+ z +….

It is seen that the formula 1), 2) and 3) fit the universal formula.

From this formula a master table can be achieved for any number of arrangements, for example :-

Number of letters | All letters different | 1 letter repeated twice | 2 letters repeated twice | 3 letters repeated twice |

4 | 24* | 12* | 6* | - |

5 | 120* | 60* | 30* | - |

6 | 720* | 360 | 180 | 90 |

7 | 5040 | 2520 | 1260 | 630 |

* shown in detail in text; the remainder direct from the universal formula.

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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