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• Level: GCSE
• Subject: Maths
• Word count: 1977

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

Extracts from this document...

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

### 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.

## 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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