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• Level: GCSE
• Subject: Maths
• Word count: 1785

# Emma Dilemma

Extracts from this document...

Introduction

Emma Dilemma

## I have been asked to look at people’s names and find the number of combinations to write them. I will look at names with 2+ letters.  I have also been asked to look at names with 2+ letters the same and find the combinations of them too.  I will try to find a pattern between them and make formulae for any of number letters.

To find out the number of combinations at the beginning I will find all the combinations with one letter at the start and then with the next letter at the front.  This is a systematic way of producing the answers and will make it much easier.  For example; if was using Lucy I would do all the combinations with L at the front and these are LUCY, LCUY, LCYU, LUYC, LYCU and LYUC I would then find the combinations with U at the beginning, these combinations are; UCLY, ULYC, UYCL, UYLC, UCYL and UCLY. As you can see this a good and simple way of finding the answer but is very long winded and would take along time to complete the longer names. So I will use a faster way to reach the answer.

Middle

20

5

6

720

360

240

30

7

5040

2520

840

210

## A name with 3 letters different

MAT, MTA

There are 2 combinations with M at the front.

2x3=6.  There are 6 combinations altogether.

3 letters with 2 letters the same

ANA, AAN

NAA

There are 3 combinations altogether, half as many as 3 different letters the same.  I can now add this data to the table.

I can work out a name with 2 letters different and a name with letters and 2 letters the same without writing them down.

I have found a pattern between names with all different letters and the names with 2 letters the same.  If you look at my table then you will b able to see that names with different letters is double names with 2 letters.  This breakthrough will help to speed up the process of working out the combinations, I can now work out the answer for all different letters and now that all I have to do is half that answer and will reach the answer for 2 letters the same.

Now I need to try and find a pattern to work out what the combinations are of whatever name with any number of letters, but first I need to do 5 letters the same and 5 letter with 2 the same.

5 letters with 2 letters the same

SCOTT, SCTOT, SCTTO, SOCTT, SOTCT, SOTTC, STOCT, STCOT, STTOC, STTCO, STOTC, STCTO

12X5=60

There are 60 combinations.

Conclusion

After working out all the formulas for words with all different letters, 2 letters, 3 letters and 4 letters the same I have spotted a pattern between their formulas.  To show this pattern if you look at the formulae for a name 2 letters the same, which is n! / 2 you have to divide n! by the number of letters the same factorial. So you could write it n! / 2! which is the same as n! / (2x1) This pattern works for every formulae.  For a name with 3 letters the same the formulae is n! / 6 which is the same as n! / 3! Which is the same as n! / (3x2x1).  This will help with the overall formulae. If we call the bottom factorial x. The formulae becomes n! / x! x= the number of letters the same (so for 2 letters the same it would be 2! and for 3 letters he same it would be 3! etc…).

The overall formula is n! / x!.

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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# Related GCSE Emma's Dilemma essays

1. ## Arrangements for names.

Total Letters (all different) Number of Arrangements 1 1 2 2 3 6 4 24 5 120 6 720 So now that I've explained the pattern of general x lettered words, what do I do if there are repeat letters?

2. ## Emma's Dilemma

I have no evidence from previous questions to determine how this factor will affect the number of arrangements, so this will be the hardest section of this question to get precise results for. This is due to the face I have no rule for working the total number of arrangements

1. ## Emma's Dilemma

a repeat I divided by a factorial, depending on the number of repeats. For example a four lettered word with 2 repeats (AABC) will be: 4! / 2! As there are two letters repeated in this case (AABB) I decided to add on another 2!. I didn't add the 2!

2. ## EMMA'S DILEMMA

ABACBC 38. BABCBC 68. CACBCB 9. ABACCB 39. BABCCB 69. CACBBC 10. ABBACC 40. BAABCC 70. CAACBB 11. ABBCAC 41. BAACBC 71. CAABCB 12. ABBCCA 42. BAACCB 72. CAABBC 13. ABCABC 43. BACBAC 73. CABCAB 14. ABCACB 44. BACBCA 74. CABCBA 15. ABCBAC 45. BACABC 75. CABACB 16. ABCBCA 46. BACACB 76. CABABC 17. ABCCAB 47. BACCBA 77.

1. ## GCSE Mathematics: Emma's Dilemma

11222112 11122212 11212221 11222121 11122221 11221122 11222211 12111222 12122112 12212121 12121212 12112122 12122121 12212211 12211212 12112212 12122211 12221112 12112221 12211122 12221121 12121122 12211221 12221211 12121221 12212112 12222111 Total Of arrangements is 70, it works The formula is confirmed This formula can be written as: a=n!/x!x!

2. ## Emma's dilemma

ADBCE 15) ADCBE 16) ADCEB 17) ADECB 18) ADEBC 19) AEBCD 20) AEBDC 21) AEDBC 22) AEDCB 23) AECDB 24) AECBD There are 24 arrangements starting just with one letter "A". therefore there are five letters altogether, so 5*24=120 Consequently there are 120 arrangements altogether. 51 = 5*4*3*2*1 = 120 arrangements.

1. ## Emma's Dilemma

the total number of letters by the previous number of combinations it was the same as multiplying the total number letters by its previous consecutive numbers Factorial Notation. The formula to find out the total number of combinations for any name when all the letters are the same is: Number

2. ## Emma's Dilemma

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