# Emma's Dilemma

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Introduction

GCSE Mathematics Coursework

Emma’s Dilemma

In my investigation I am going to investigate the number of different arrangements of letters for names and words and try to find a formula that can be used to predict this.

For example: TOM is one arrangement

and OTM is another arrangement

First, I am going to investigate the number of different arrangements of letters for the name LUCY (a 4-letter name, where all the letters are different).

LUCY ULCY CLUY YLUC

LUYC ULYC CLYU YLCU

LCUY UCLY CULY YULC

LCYU UCYL CUYL YUCL

LYUC UYLC CYLU YCLU

LYCU UYCL CYUL YCUL

There are 4 different letters and 24 different arrangements.

Once I have investigated the number of different arrangements for one 4-letter name/word where all the letters are different, I do not need to try any more. If I tried the name DAVE for example, there would still be 24 different arrangements. I could substitute the L in LUCY for the D in DAVE, the U for A, the C for V, and the Y for E; and would therefore end up with the same result. The same is true for names/words with 3 letters or 5 letters, etc. As long as the number of letters and the number of different letters are the same, the number of different arrangements will be the same.

Now I will investigate a 3-letter name where all the letters are different.

SAM ASM MSA

SMA AMS MAS

There are 6 different arrangements.

Now I will investigate a 2-letter name where all the letters are different.

JO OJ

There are 2 different arrangements.

Now I am going to investigate a 5-letter name where all the letters are different.

KATIE AKTIE TKAIE IKATE EKATI

KATEI AKTEI TKAEI IKAET EKAIT

KAITE AKITE TKIAE IKTAE EKTAI

KAIET AKIET TKIEA IKTEA EKTIA

KAETI AKETI TKEAI IKEAT EKIAT

KAEIT AKEIT TKEIA IKETA EKITA

KTAIE ATKIE TAKIE IAKTE EAKTI

KTAEI ATKEI TAKEI IAKET EAKIT

Middle

24 ÷ 2 = 12 different arrangements.

Why you divide by 2:

When finding the number of different arrangements for words that have one letter repeated twice, you halve the number of different arrangements for a word which has the same amount of letters, but none repeated.

If we look at EMMA (a 4-letter word) for example, it has the same letter (M) repeated twice, and has 12 different arrangements. LUCY (a 4-letter word with no letters repeated) has 24 arrangements.

If we take each M in EMMA to be a different letter, by changing the colour of one of the Ms, we would get 24 arrangements.

EMMA MEMA MEMA AEMM

EMAM MEAM MEAM AEMM

EMMA MMEA MMEA AMEM

EMAMMMAE MMAE AMME

EAMM MAEM MAEM AMEM

EAMMMAME MAME AMME

But for this investigation EMMA and EMMA are the same arrangement because M and M are the same letter (M) and make no difference to the arrangement; they are both the arrangement EMMA.

EMMA and EMMA are both EMMA.

EMAM and EMAM are both EMAM.

EAMM and EAMM are both EAMM.

MEMA and MEMA are both MEMA.

MEAM and MEAM are both MEAM.

MMEA and MMEA are both MMEA.

MMAE and MMAE are both MMAE.

MAEM and MAEM are both MAEM.

MAME and MAME are both MAME.

AEMM and AEMM are both AEMM.

AMEM and AMEM are both AMEM.

AMME and AMME are both AMME.

If each M were different, then for each arrangement there would be another similar arrangement because you could just swap the Ms around (e.g. EMMA → swap Ms around → EMMA), so there are twice the number of arrangements. But since the Ms are the same for the purpose of this investigation, there is only half the number of different arrangements as there would be if the 2 Ms were different letters, so you divide by 2.

Conclusion

XXYY XYXY XYYX YXYX YYXX YXXY

XXYY XYXY XYYX YXYX YYXX YXXY

But because another letter is repeated (Y), for each of the 12 arrangements the Ys can also be rearranged. So you have to multiply the number of different ways the 2 Ys (if they were different) can be arranged (found by using n! = a) by 12 to get 24 different arrangements.

XXYY XYXY XYYX YXYX YYXX YXXY

XXYYXYXYXYYX YXYX YYXX YXXY

XXYY XYXY XYYXYXYXYYXXYXXY

XXYY XYXY XYYX YXYX YYXX YXXY

This is why you have to multiply the number of repeats factorial for each different letter together.

The formula for finding the number of different arrangements of letters for any words/names is:

n! = a

x! y!

## Where

n = the total number of letters in the word/name

x = the number of times one letter is repeated in the word/name

y = the number of times another letter is repeated in the word/name

a = the number of different arrangements

Using this formula, I can predict the number of different arrangements of letters for any word/name.

I predict that the number of different arrangements for the word XXYYZ is:

n! = a 5! = 30

x! y! z! 2! × 2! × 1!

XXYYZ XXYZY XXZYY XYXYZ XYXZY XYYXZ

XYYZX XYZXY XYZYX XZXYY XZYXY XZYYX

YXXYZ YXXZY YXYXZ YXYZX YXZXY YXZYX

YYXXZ YYXZX YYZXX YZXXY YZXYX YZYXX

ZXXYY ZXYXY ZXYYX ZYXXY ZYXYX ZYYXX

There are 30 different arrangements.

I predict that the number of different arrangements for the word WWXYZ is:

n! = a 5! = 60

w! x! y! z! 2! × 1! × 1! × 1!

WWXYZ WXZWY WZXWY XYWWZ YWZWX ZWXWY

WWXZY WXZYW WZXYW XYWZW YWZXW ZWXYW

WWYXZ WYWXZ WZYWX XYZWW YXWWZ ZWYWX

WWYZX WYWZX WZYXW XZWWY YXWZW ZWYXW

WWZXY WYXWA XWWYZ XZWYW YXZWW ZXWWY

WWZYX WYXZW XWWZY XZYWW YZWWX ZXWYW

WXWYZ WYZWX XWYWZ YWWXZ YZWXW ZXYWW

WXWZY WYZXW XWYZW YWWZX YZXWW ZYWWX

WXYWZ WZWXY XWZWY YWXWZ ZWWXY ZYWXW

WXYZW WZWYX XWZYW YWXZW ZWWYX ZYXWW

There are 60 different arrangements.

I have found a formula that can be used to find the number of different arrangements of letters for any word or name, explained and proved why it works, and have achieved the aim of my investigation.

By Thomas Tam 10B1

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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