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  • Level: GCSE
  • Subject: Maths
  • Word count: 2883

GCSE Mathematics: Emma's Dilemma

Extracts from this document...

Introduction

GCSE Mathematics: Emma's Dilemma Emma and Lucy are playing with arrangements of their names. One arrangement of LUCY is L.U.C.Y. Another arrangement is Y.L.C.U. We are investigating the number of different arrangements of Emma and Lucy's names. Then we shall move onto the number of different arrangements of various groups of letters. Firstly, I shall list all the possible combinations for the word "LUCY": LUCY LCUY LYUC LYCU LUYC LCYU UCYL UYCL ULCY ULYC UYLC UCLY CYLU CYUL CULY CUYL CLYU CLUY YCUL YCLU YUCL YULC YLUC YLCU = 24 Combinations I listed all these combinations by using the following method: Step 1 Arrange the listing process into 4 stages: 1 _ _ _ 2 _ _ _ 3 _ _ _ 4 _ _ _ Step 2 You start off with the original word: 1234 Acquire the combinations of the last two numbers first and you end up with 1243. Now that you have 1243, do the last three numbers and try the different possibilities: 1423 1432 1342 1324. Because the number 2 has been the first number of last three numbers, we don't do it again. Step 3 Now that we have listed all the arrangements beginning with 1, we do the list of arrangements with 2 in the beginning: Start off with 2134 and do same thing to it, it will look like this: 2134 2143 2431 2413 2314 2341 Step 4 Now that we have established the different arrangements of 2, we go ahead and do 3 at the beginning: 3124 3142 3241 3214 3412 3421 Step 5 Now that we have established the different arrangements of 3, we go ahead and do 4 at ...read more.

Middle

Try 5 letters: 42213 12234 42231 12243 42123 12324 42132 12342 42321 12423 42312 12432 41223 13224 41232 13242 41322 13422 43122 14223 43212 14232 43221 14322 21234 23124 21243 23142 21324 23214 21342 23241 21423 23421 21421 23412 22134 24123 22143 24132 32412 22314 24231 32421 22341 24213 34122 22413 24312 34212 22431 24321 34221 32241 32214 32142 32124 31422 31242 31224 So the total arrangements are 12 x 5=60 Let's put our results in a table and try to extract a formula from there: Number Of Letters (With 2 of the letters the same) Number Of Combinations 2 1 3 6 4 12 5 60 Which can be re-written as: Number Of Letters (With 2 letters the same) Number Of Combinations 2 1 3 3 x 1 4 4 x 3 x 1 5 5 x 4 x 3 x 1 Let's work out the formula: if n= number of figures a= number of arrangements the formula is a=n!/2 Let's confirm the formula: Number Of Letters (with 2 letters the same) Number Of Combinations 2 2/2=1 It works 3 (1 x 2 x 3) / 2=3 It works 4 (1 x 2 x 3 x 4) / 2=12 It works Now I shall investigate different amounts of letters with different amounts of repeated letters. I shall start off with a four letter word where 3 of the letters are the same. For this I shall use the combination "3331": 3331 3313 3133 1333 - There are 4 combinations. Now let's try a 5 letter word with 3 letters the same: 33312 31233 33321 31323 33123 31332 33132 32331 33231 32313 33213 32133 21333 12333 23133 13233 23313 13323 ...read more.

Conclusion

The formula is confirmed. Conclusion From the investigation above we find out the formula for calculating the number of arrangements, it's a=n!/x! a represent the total arrangements n represent the number of figures of the number ! represents the factorial function. x represent the number of duplicated letters in the word for 2 pairs of the same letter: the formula is a=n!/x!x! for 2 pairs of different duplicated letters of same number in a word: the formula is a=n!/x1!x2! for 3 pairs of same number of duplicated letters in a word: the formula is a=n!/x!x!x! for 3 pairs of a different number of the same letter in a word: The formula is a=n!/x1!x2!x3!. The formula can be also used to find the amount of different arrangements of letters. For example: xxyy the arrangement for this is a=(4 x 3 x 2 x 1)/(2 x 1 x 2 x 1)=6 xxyyy the arrangement for this is a=(5 x 4 x 3 x 2 x 1)/(3 x 2 x 1 x 2 x 1)=10 xxxxxxyyyyyyyyyy The formula for this is: a=n!/x1!x2!=(16 x 15 x 14 x 13 x 12 x 11 x 10 x 9x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ) / ( 1 0 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 3 x 2 x 1 x 6 x 5 x 4 x 3 x 2 x 1 ) = 8 0 0 8 The total arrangement is 8008. Using this formula, we can find out the total arrangements of all numbers and letters. Sheroy Zaq (X2) Mrs Whybrow ...read more.

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