# Emma is playing with arrangements of the letters of her name to see how many combinations there are.

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Introduction

Year 11 Coursework:

## Emma’s Dilemma

Emma is playing with arrangements of the letters of her name to see how many combinations there are.

EMMA AEMM MMAE MAEM

EMAM AMEM MMEA MEMA

EAMM AMME MEAM MAME

I found that there are 12 combinations of the letters in the name Emma. However, I think that the fact that there is a repeated letter in the name has affected the result. I will try another name with no repeated letter to see if it will affect the number of combinations. I will use the name Lucy.

## LUCY UCYL CYLU YLUC

LUYC UCLY CYUL YLCU

LCUY ULCY CUYL YCUL

LCYU ULYC CULY YCLU

LYUC UYLC CLYU YUCL

LYCU UYCL CLUY YULC

I found 24 different combinations, twice as many for Emma. I will try another couple of four-letter names to see if I get the same results. I will try another name with a repeated letter to see if I find 12 combinations.

## NINA NANI AINN IANN

NIAN NNAI ANIN INAN

NAIN NNIA ANNI INNA

Like Emma, there are 12 combinations for the name Nina.

To make sure I don’t miss out any combinations or repeat any I use a system.

I start with the name then swap around the last two letters.

Eg. NINA

## NIAN

I then change the second letter and repeat this.

NANI

NAIN

Middle

KRYER EKYRR REKYR RKRYE

YKERR YERRK YRREK YREKR

YKRER YERKR YRRKE YRKRE

YKRRE YEKRR YRKER YRERK

There are 60 combinations for a five-letter name with one repeated letter.

I will now try a name with two repeated letters I will use the name Hanna.

HANNA ANNAH NANAH

HANAN ANNHA NANHA

HAANN ANHNA NAHNA

HNNAA ANHAN NAHAN

HNANA ANANH NAAHN

HNAAN ANAHN NAANH

AAHNN NNAAH

AANHN NNAHA

AANNH NNHAA

AHNNA NHAAN

AHNAN NHANA

AHANN NHNAA

I have found 30 different combinations for a five-letter name with two repeats.

No Repeats | 1 Repeat | 2 Repeats | |

5 Letters | 120 | 60 | 30 |

I can see from my table that the pattern for the results for five-letter names is the same. The number of arrangements is halved each time a letter is repeated.

I am trying to find a formula to find the number of combinations for any given number of letters. I will find results for six-letter names to help me find the formula. I predict that there will be 720 combinations for a six-letter name with no repeats. I think this because there are 6 combinations beginning with NIC;

## NICOLA NICALO NICLOA

### NICOAL NICAOL NICLAO

### Therefore there will be 6 combinations each for NIL, NIA and NIO.

Eg. NILCAO NILACO NILOCA

NILCOA NILAOC NILOAC etc.

There will therefore be 24 combinations for NI, 24 for NC and so on. Altogether there will be 120 combinations beginning with the letter N.

Conclusion

No. of x’s

No. of combinations

1

1! (= 1)

1

1

2

2! (= 2)

2

1

3

3! (= 3)

3

1

4

4! (= 4)

4

1

I know that n! equals the number of combinations for any word with no

x!

repeated letters and n! is for one repeated letter etc. However, I want to

2!

find out the formula for working out the number of combinations for words with letters repeated more than once.

Eg. XXXYYY

The formula for this could be written as:

6! (no.of letters)

3! × 3! (no. of x’s > y’s)

Or

720 = 20

36

I will write out all thepossiblecombinations for XXXYYY to check my result.

## XXXYYY

### XXYXYY

## XXYYXY

XXYYYX

XYXXYY

XYXYXY

XYXYYX

XYYXXY

XYYXYX

XYYYXX

YXXXYY

YXXYXY

YXYXXY

YXYXYX

YXYYXX

YYXXXY

YYXXYX

YYXYXX

YYYXXX

YXXYYX

My formula worked correctly – I have found 20 possible combinations for XXXYYY.

I can now do a formula for any word with any number of letters and any number of repeats.

For example:

For XXYYZZ it would be

6!

2! 2! 2!

For XXYZABC it wouldbe

7!

2! 1! 1! 1! 1! 1!

My overall formula would be:

n!

a!b!c!d!e!f!g!h!i!j!k!l!m!n!o!p!q!r!s!t!u!v!w!x!y!z!

This can be used to find out the number of combinations for the letters of any word. I would just have to substitute the number of each letter for the letter. So, instead of a!, I would put in the number of times that a appears in the word I was trying to find out the number of combinations for.

For example:

‘Maths’ would be:

5!

1! 1! 1! 1! 1!

Which would equal 120 different combinations.

Whereas ‘classes’ would be:

7!

1! 1! 1! 3! 1!

Which would equal 840 different combinations.

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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