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• Level: GCSE
• Subject: Maths
• Word count: 1465

# Find out how many different arrangements of letters there are in a name or word.

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Introduction

EMMA’S DILEMMA

In this piece of coursework, I am going to find out how many different arrangements of letters there are in a name or word (according to the number of letters). I will then try to find a pattern in the arrangements that will help me to form an equation. I will also try to find out how many different arrangements there are in a word with two or more sets of different letters that occur more than once.  E.g. Xxyyaa .I know this sounds very confusing but you’ll soon understand what I’m talking about. By the end of this piece of course work, I should be able to find out how many arrangements there are in XXXYYVVVV by using an equation.

1) My first name will be EMMA:

 1)EMMA 2)AMME 3)MMEA 4)MEAM 5)EAMM 6)EMAM 7)AMEM 8)AEMM 9)MAEM 10)MMAE 11)MEMA 12)MAME

I have concluded that there are 12 different arrangements in this name. I used various methods to find out the different arrangements: I would put the letter at the beginning on the end each time. E.g.

Middle

12)CSEAY

13)CESAY

14)CESYA

15)CEYAS

16)CEYSA

17)CEAYS

18)CEASY

19)CYAES

20)CYASE

21)CYESA

22)CYEAS

23)CYSEA

24)CYSAE

As you can see I haven’t written out all the arrangements for CASEY because this is not necessary. I found all the possible arrangements beginning with C and have multiplied it by the number of letters in the name which is 5.

So, 5 ÷ 24=120. This means there are 120 different arrangements in the name CASEY.

I think I have discovered a pattern:

 Name No. of letters No. of arrangements MO 2 2 RIA 3 6 LUCY 4 24 CASEY 5 120

If you look at MO, the number of arrangements is 2. For RIA the number of arrangements are 6 and for LUCY, 24. If you multiply the number of letters by the number of arrangements previously, then you will get the number of arrangements for that number of letter. Sounds confusing, here are a couple of examples.

No. of letters × no. of arrangements previously = no. of arrangements

RIA:                       3           ×                          2                        =              6

LUCY:             4           ×                          6                        =             24

CASEY:           5           ×                         24                       =           120

I can now use this equation to find out arrangements of other words with more letters.

 Name No. of letters No. of arrangements MAXINE 6 720 RIZWANE 7 5040 SAMENTHI 8 40320

To make this easier you could say:

1 × 2= 2 = 2!

1 × 2 × 3= 6 = 3!

1 × 2 × 3 × 4= 24= 4!

1 × 2 × 3 × 4 × 5= 120 = 5!

1 × 2 × 3 × 4 × 5 × 6= 720 = 6!

1 × 2 × 3 × 4 × 5 × 6 × 7= 5040 = 7!

1 × 2 × 3 × 4 × 5 × 6 × 7 × 8= 40320 = 8!

I will now see what happens when you have a name or word that has a letter in it that occurs twice or more and compare it to the same letters of a word where none of the letters repeat.

4 letters:

 Name/Word No. of times the same letter occurs No. of arrangements LUCY 0 24 EMMA 2 12 ANNN 3 4 NNNN 4 1

Conclusion

No. of times B occurs

No of arrangements

ABBA

4

2

2

6

AAABB

5

3

2

10

ABBA: There are 4 letters in the word so I at first thought it might be 4! But this cannot be because there are two sets of different letters which occur more than once. I thought that there must be fewer arrangements because of this. So I thought:

ABBA = 4! ÷ (2! + 2!) = 6. This equation worked so I tried:

AAABB = 5! ÷ (3! + 2!) = 15. But this wasn’t right as there are only 10 arrangements.

At last I tried: 5! ÷ (3! × 2!)= 10. This worked. I tried it again with ABBA and it worked. Now I have found the equation:

No. of letters in factorial ÷ (No. of times a letter repeats × No. of times a letter repeats) = no. of arrangements

At the beginning of the course work I said that I wanted to be able to work out how many different arrangements there are in XXXYYVVVV by the end of the coursework. I can now do that: Arrangements =

η!                         9!               362880          362880

(χ! × γ! × ν!)     (3! × 2! × 4!)    (6 × 2 × 24)        288

η! = 9! = no. of letters in the word

χ! = 3! = no. of x in the word

γ! = 2! = no. of Y’s in the word

ν! = 4! = no. of V’s in the word

Hope you enjoyed this thoroughly interesting piece of course work and I look forward to a good grade (PLEASE?) Bye!

By RIA Hylton 10A

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