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  • Level: GCSE
  • Subject: Maths
  • Word count: 1107

Emma's dilemma.

Extracts from this document...

Introduction

Investigation

To start my investigation I am going to start with something a little easier than Lucy this will help with trying to establish a general pattern and in turn the general formula.

I will start with the word IT

Name

Combinations

Number of letters

IT

IT

2

TI

This is very simple the word has 2 letters and 2 possible arrangements for the letters.

Now I will try IAN

Name

Combinations

Number of letters

IAN

IAN

3

INA

NIA

NAI

AIN

ANI

The number of letters that I have used has now increased to 3 but the number of arrangements has changed it has become 6.


...read more.

Middle

Lucy 4x3x2x1= 24 (number of combinations)

I can best describe how I came to this theory by using an example involving cards and spaces. If I have 4 cards and four spaces and I place one of the cards in one of the spaces I have three cards and three spaces. If I put another card  down I have two cards left along with two spaces if I do this again I have one remaining card and one space. If I fill this space then I have no cards and no spaces. This is the same with the letters for example of Lucy’s name. I found that by multiplying the number of places and letters there were to fill the places I could get the total number of arrangements for the word. This is called factorial notation and is symbolised by an !

Factorial notation is multiplying the total number of letters by the previous consecutive numbers giving the total number of positions and therefore arrangements of the words.

Therefore the formula for total number of combinations for a word with no repeats is C=T!

...read more.

Conclusion

My new formula is correct

Why my formula works

The number of repeats is made factorial because it shows how many combinations will be made the same by the repeats. Dividing by this gets rid of the combinations that are the same because of the repeats.

When you have more that one set of repeats you multiply them together to get the total number of combinations that need to be removed regardless of letter. This is why my formula works.


Conclusion

  • To find the total amount of arrangements for a letter with no repeats you use it’s amount of letters and make it factorial.
  • The formula for this is C=T!
  • As the number of letters increases so does the amount of arrangements for these letters.
  • When there are repeats in the words the number of different combination decreases.
  • A new formula is created to allow for the repeats and by division removes them from the equation. This formula is C=T!/A!B!C!
  • The total number of repeated letters are made factorial and multiplied together to give a number that is used in the formula to give the final answer.
  • This formula works with all cases regardless of how many letters there are or how many repeats there are this will always work.

...read more.

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