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  • Level: GCSE
  • Subject: Maths
  • Word count: 4324

Emma's Dilemma

Extracts from this document...

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

...read more.

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.

...read more.

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

...read more.

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