# 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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