# Emma&#146;s dilemma

Extracts from this document...

Introduction

Emma’s dilemma

Every name has many ways of arranging the letters e.g. BOB, OBB, BBO. I am trying to find a pattern and formula for finding out how many combinations there for a name. I tried different names with a different number of letters in them, I also tried the names with repeated letters in them. There are several methods of finding out the combinations of a name, I have used 3 different methods.

## Method 1

## Listing

## Two lettered names

ED

DE

2 ways

Three lettered names

As you can see below a three lettered name with 1 repetion in it has half the number of combinations.

TED

TDE

ETD

EDT

DTE

DET

6 ways

BOB

BBO

OBB

3 ways

Four lettered names

I have noticed that with four lettered names that with 1 letter repeated there

Middle

L 24 E24 O 4

A 24 T12 Y 4

R 24 R12 20 WAYS

K 24 60 WAYS

## 120 WAYS

## 6 lettered names

ROBERT WILLIE THOMAS

R120 W30 T120

O60 I 60 H120

B60 L60 O120

E60 E30 M120

T60 180 WAYS A120

## 360 WAYS S120

720 WAYS

7 LETTERED NAMES

WILLIAM PHILLIP

W180 P 180

I 360 H 90

L 360 I 180

A 180 L 180

### M 180 630 WAYS

## 1260 WAYS

## Method 2

Probability

If there was a bag with 3 pieces of paper in it 1 with t, 1 with e and 1 with d the chances of getting the letters in order is 1/6 because the chances of getting t the first time is 1/3, if you do get the t then there is only 2 pieces left in the bag so getting e the first time

Conclusion

2/7 x 1/6 x 2/5 x 2/4 x 1/3 x 1/2 x 1/1

### Method 3

## Factorial

! is the symbol in maths for factorial, this means the number multiplied by all the numbers before it up to 0 e.g. 5! Would be 5x4x3x2x1.

2 LETTERED NAMES

ED

½! = 2 WAYS

3 LETTERED NAMES

TED

1/3! =3 WAYS

BOB

2!/3!= 2/6 =1/3 = 3 WAYS

4 LETTERED NAMES

LUCY

¼!= 24 WAYS

EMMA

2!/4!= 2/24 = 12 WAYS

LULU

2!x2!/4!= 4/24 =6 ways

5 LETTERED NAMES

CLARK

1/5!=120 WAYS

PETER

2!/5!=60 WAYS

BOBBY

3!x2!/5!= 20 WAYS

6 LETTERED NAMES

ROBERT

2!/6!=360 WAYS

WILLIE

2!x2!/6!=180WAYS

THOMAS

1/6!= 720 WAYS

7 LETTERED NAMES

WILLIAM

2!x2!/7!=1260 WAYS

PHILLIP

2!x2!x2!/7!= 630 ways

From all the lists I have made I was able to find the formula

E.g. 5 3

xxxxx yyy the equation for this would be (5+3)!/5!x 3!

So if there are x amount of Xs and y amount of Ys e.g.

X y

(xxxx…..x)(yyyy….y) the formula would be (x+y)!/X! x Y!

The final formula to work out how many combinations there are in a name is

(X+Y)!

~~~~~~

X! x Y!

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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