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Emma’s dilemma

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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

...read more.

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

...read more.

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!

...read more.

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