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• Level: GCSE
• Subject: Maths
• Word count: 1378

# Emma&amp;#146;s Dilemma

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Introduction

Emma's Dilemma Aim: The aim of this investigation is to study the number of combinations, which can be used for different words. I will examine words where the letters are all different or where there may be more than one repeated letter. I will examine words with different numbers of letters. I will then eventually work out a formula for it. Investigation: In the name LUCY there are four different letters and there are 24 different arrangements: 1. 1. LUCY 2. LUYC 3. LCUY 4. LCYU 5. LYCU 6. LYUC 7. ULYC 8. ULCY 9. UCLY 10. UCYL 11. UYCL 12. UYLC 13. CLUY 14. CLYU 15. CYUL 16. CYLU 17. CUYL 18. CULY 19. YLUC 20. YLCU 21. YCLU 22. YCUL 23. YUCL 24. YULC In the three-letter word SAM, there are 3 different letters and 6 different arrangements: 1. 1. SAM 2. SMA 3. MAS 4. MSA 5. ASM 6. AMS In the two-letter word, MO there are 2 different letters and 2 different arrangements: 1. 1. MO 2. OM Number of all different Letters Number of Different Arrangements 1 1 2 2 3 6 4 24 5 120 6 720 7 5040 If we put all of this data into a table; From the above table of results, I have found out that a two-letter word has two arrangements, and a three-letter word has six arrangements. ...read more.

Middle

0 5 120 60 20 5 1 6 720 360 120 30 6 7 5040 2520 840 210 42 Let us look for a formula if a word has two same letters such as EMMA. If we study the table we can work out that if n= no of letters and a=no of arrangements, the formula is a=n! ? 2. Let us prove this formula for the name EMMA. If we use the formula, it is a four-letter word so a=n! ? 2 = 4! ? 2= 12. If we look above at the table and the list below we can see that EMMA does have 12 arrangements: 1. MMAE 2. AMME 3. AMEM 4. EMAM 5. AEMM 6. EAMM 1. 7. MMEA 8. MMAE 9. MEMA 10. MAME 11. MEAM 12. MAEM Let us look at one more word. BAA is a three-letter word with two repeated letters. If we feed it into the formula; a=n! ? 2= 3! ? 2= 3. If we look at the table above and the list below we can see that our formula is confirmed: 1. BAA 2. ABA 3. AAB Now let us find a formula with a word with three letters the same. If we study the table can work out that if n= no of letters and a= no of arrangements the formula is a= n! ...read more.

Conclusion

YYXX XXXYY is a five-letter word with two letters the same (3X's and 3Y's). There are ten different arrangements: 1. XXXYY 2. XXYXY 3. XXYYX 4. XYXYX 5. XYXXY 6. XYYXX 7. YYXXX 8. YXXXY 9. YXYXX 10. YXXYX XXXYY has 3X's and 2Y's. As each letter has its own quantity of arrangements, for example in XXXYY there were 6 arrangements beginning with x, and 4 arrangements beginning with y, I think that the factorial has to be used again in the formula. I have come up with a new formula. The number of total letters (n) factorial, divided by the number of X's and Y's factorised and multiplied together. If n! = the number of letters in a word and x! y! = the number of repeated letters the same and a= the number of arrangements I came up with the formula: a= n! (x!)(y!) If you test this formula on the previous examples; XXYY has four letters and 2X's and 2Y's. So: a= n! = 4! = (4? 3? 2? 1) = 6 (x!) (y!) (2!) (2!) (2?1) (2?1) If we look at the list on the previous page, we can see se that XXYY does have six arrangements. If we try another example; XXXYY has five letters and 3X's and 2Y's. So: a= n! = 5! = (5? 4? 3? 2? 1) = 10 (x!) (y!) (3!) (2!) (3?1) (2?1) If we look at the list, we can see that the word XXXYY does have 10 combinations. 1 1 ...read more.

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1. ## Emma's Dilemma

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1. ## Emma's Dilemma

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cyblu ycblu bylcu lybcu uybcl cybul ycbul byulc lyubc uylbc cylbu yclbu byucl lyucb uylcb cylub yclub byclu lycbu uycbl cyubl ycubl bycul lycub uyclb cyulb yculb As you can see there are 120 different combinations for a 5-letter word.

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As you can see my method of finding all the permutations allows them to be listed in ascending numerical order without missing out any. Altogether there are 12 possible arrangements of the letters in the name 'Emma' with no permutations missing or duplicated.

2. ## Emma&amp;amp;#146;s Dilemma

This helps to make finding the number of arrangements a lot easier if you have the information about one letter. However, finding all the arrangements of one of the letters still takes time especially with a long name. To find out a quicker way I firstly looked into some other

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