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• Level: GCSE
• Subject: Maths
• Word count: 1288

# Emma&amp;#146;s Dilemma

Extracts from this document...

Introduction

Emma's Dilemma Introduction In this project I will investigate the number of different ways in which the word 'EMMA' can be spelt. I am also going to investigate the word 'LUCY' as it does not contain any double letters and I think it will result in a different answer compared to 'EMMA'. I will also back these up by working with letters 'JOHN' and 'ABBY'. I will then try a set of two double letters such as 'ABBA' and 'JOJO to see how the results differ with two double letters compared with one double letter. I will also investigate combinations of the letters 'X' and 'Y' and see the number of different arrangements I can make. I will try to be methodical in the way in which I work these out so that I do not miss any results out. For 'EMMA' I will firstly find out all the arrangements that start with 'E'. Then I would test 'M' and then 'A' Investigation EMMA In the name Emma there are 12 possible arrangements. There are three arrangements for E, three arrangements for A and six arrangements for M which I would have expected because there are two M's in the name. ...read more.

Middle

for JON there were 6 arrangements and 1 x 2 x 3 = 6. Also with JOHN the total number of outcomes was 24 and 1 x 2 x 3 x 4 = 24. PREDICTION 2: using the rule above, I am going to predict that the 5 letter name I will choose, which is without any double letters is as follows: 1 x 2 x 3 x 4 x 5 = 120 BERTY PREDICTION 2: My Prediction was correct Double Letters I am now going to investigate the number of different arrangements I can make with name with double letters in them. I will compare these results with my single letter results BB BB There is only one possible arrangement for this name BOO BOO BOB OBB I noticed that there were three outcomes for the three letter name ABBY This name has the same outcome as the name EMMA. The number of outcomes for this name is twelve names. I have noticed that it is half of the single name results. In theory if I use the same rule as before but divide that whole rule by two because there is a double letter, it should work. ...read more.

Conclusion

No. Of X 1Y 2Y 3Y 4Y 5Y 1 2 3 4 5 six 2 3 6 10 15 twenty one 3 4 10 20 fifty six fifty six 4 5 15 35 seventy one-hundred and twenty six 5 six twenty one fifty six one-hundred and twenty six two-hundred and fifty two 6 seven twenty eight eighty four two-hundred and ten four-hundred and sixty two If we go back to the beginning of the investigation and use the name EMMA, we can see that it has three different letters in it, E, M and A. if we take the rule for the double letters, (n!/2) but instead of dividing by two, you divide by the how many different letters there are, you end up with the total number of outcome that are possible. E.g. 4! (1!x2! x1!) So if we want to find the entire possible outcome for HOMER, we would have to write 5! (1!x1! x1! x1! x1!) this gives us the answer of 120, which is the answer to the 5 single letters. To find out the X's and Y's, we adapt the equation a little bit and put n!/X! x Y! . X means total number of X's and Y mean total number of Y's and this is all divided by the total number of letters to find the X's and Y's. ...read more.

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1. ## Emma's Dilemma

But we still follow the same procedures. There are four starting letters (A,B,C and D), and three letters which go with them. And from our last case we already know that with three letters, there are 6 combinations (3x2x1). So now, we just multiply this by 4 4x3x2x1 (or 4 multiplied by 6)

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

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.

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a relation between the amounts by which the arrangements increased for words with the same number of letters: 1) one repeated letter 2) words with no repeated letters. So I added an extra row onto the table and tried to find a link between the two.

2. ## Emma's Dilemma

1 Name: "BBBC" Number of letters: 4 Arrangements: BBBC BCBB BBCB CBBB Total Number of Arrangements: 4 Name: "CCCDE" Number of letters: 5 Arrangements: CCCDE CCDCE CECDC DECCC CCCED CDCCE CEDCC ECCCD CCEDC CDCEC DCCCE ECCDC CCECD CDECC DCCEC ECDCC CCDEC CECCD DCECC EDECC Total Number of Arrangements: 20 Hypothesis:

1. ## Emma's Dilemma

ELIAUP ALIUAP ELIUPA EIPAUL EIPALU EIPLAU EIPLUA EIPULA EIPUAL EIAPUL EIPLU EIALPU EIALUP EIAULP EIAUPL EIUPAL EIUPLA EIULPA EIULAP EIUALP EIUAPL EILPAU EILPUA EILUPA EILUAP EILAPU EILAUP = 720 So the total number of arrangement for a six-lettered word whose letters are all different is 720.

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120 divided by 6 (number of letters) equals 24. Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter. For the sake of convenience I have used lucy and put a q in front of it to show that there

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2. ## Emma's Dilemma

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