Emma’s Dilemma

I will now investigate the number of different arrangements the name JO can give.

JO

JO

OJ

The name JO only gave two different arrangements, as there is only one option after the first letter. Then it is reversed to make the second arrangement e.g. JO/OJ

I will now investigate the number of different arrangements the name JON can give

JON

When I did the name JON, I noticed that the first letter had two letter options after it and that was two was the only number of arrangements for that starting letter e.g. J-ON / J-NO

PREDICTION 1: I predict that when I do four letters, there will be three outcomes for the starting letter.

JOHN

PREDICTION 1: FAILED

The prediction I made was not correct. This was because the Starting letter has three options after it, then the second letter has two options after it. This means it will have the same amount of options as JON but with an extra option on top of it H. If the letters were given numbers, it would be easier to see. In the name JON you would have 123 and in the name JOHN you would have 1234. If we multiply the numbers together we get the total number of arrangements that can be made

E.g. for JON there were 6 arrangements and 1 x 2 x 3 = 6. Also with JOHN the total number of outcomes was 24 and 1 x 2 x 3 x 4 = 24.

PREDICTION 2: using the rule above, I am going to predict that the 5 letter name I will choose, which is without any double letters is as follows: 1 x 2 x 3 x 4 x 5 = 120

BERTY

PREDICTION 2: My Prediction was correct

Double Letters

I am now going to investigate the number of different arrangements I can make with name with double letters in them. I will compare these results with my single letter results

BB

BB

There is only one possible arrangement for this name

BOO

BOO

BOB

OBB

I noticed that there were three outcomes for the three letter name

ABBY

This name has the same outcome as the name EMMA. The number of outcomes for this name is twelve names. I have noticed that it is half of the single name results. In theory if I use the same rule as before but divide that whole rule by two because there is a double letter, it should work. E.g. 1 x 2 x 3 x 4 = 12

2

PREDICTION 3: I predict for the 5 letters, the total would be

1 x 2 x 3 x 4 x 5 = 60

2

DAVID

PREDICTION: My Prediction was correct

Triple Letters

Now I will investigate names with triple letters inside them. I will substitute the names for random letters, as there are not ant triple letter names I can think of.

AAA

AAA

BAAA

BAAA

ABAA

AABA

AAAB

I have noticed that these results can be found by dividing the number of letters factorial by 6.

PREDICTION 3: I predict that the next set of letters will be...

E.G. 5!

2

BAAAH

PREDICTION 3: My prediction is correct.

Table of Results

No. Of Letters

Single Letters

Double Letters

Triple Letters

~-~

~-~

2

2

~-~

3

6

3

4

24

2

4

5

20

60

20

Rules I have found

Single Letters: n!

Double Letters: n!

2

Triple Letters: n!

6

X's And Y's

Now I will investigate the number of different arrangements I can make with the letters X and Y. I will try and find a master rule for this.

XY

XY

YX

XXY

XXY

XYX

YXX

YYX

YYX

YXY

XYY

XXYY

XXYY

XYXY

XYYX

YYXX

YXYX

YXXY

XXXY

XXXY

XXYX

XYXX

YXXX

XXXYY

XXXYYY

XXXXY

PREDICTION 4: There is a pattern. If you have 6X's and 1Y, there will be 5 arrangements. I predict that there will be 6 arrangements for 5X's and 1Y.

XXXXXY

PREDICTION 4: My prediction was correct.

XXXXYY

PREDICTION 5: There is a pattern when there are 2Y's. Each X increases the number of arrangements in triangular numbers. I predict that there will be 21 arrangements for the next name.

XXXXXOO

PREDICTION 5: My prediction was correct, I predict the next one will be 35

XXXXYYY

Table of Results (X's and Y's)

No. Of X

Y

2Y

3Y

4Y

5Y

2

3

4

5

six

2

3

6

0

5

twenty one

3

4

0

20

fifty six

fifty six

4

5

5

35

seventy

one-hundred and twenty six

5

six

twenty one

fifty six

one-hundred and twenty six

two-hundred and fifty two

6

seven

twenty eight

eighty four

two-hundred and ten

four-hundred and sixty two

If we go back to the beginning of the investigation and use the name EMMA, we can see that it has three different letters in it, E, M and A. if we take the rule for the double letters, (n!/2) but instead of dividing by two, you divide by the how many different letters there are, you end up with the total number of outcome that are possible. E.g. 4! (1!x2! x1!)

So if we want to find the entire possible outcome for HOMER, we would have to write

5!

(1!x1! x1! x1! x1!)

this gives us the answer of 120, which is the answer to the 5 single letters. To find out the X's and Y's, we adapt the equation a little bit and put n!/X! x Y! . X means total number of X's and Y mean total number of Y's and this is all divided by the total number of letters to find the X's and Y's.