From Emma's name, a four letter word, we this time only get twelve arrangements. This exactly half of twenty-four,(twenty-four being the total number of arrangements for a four letter word with all the letters differ

        LUCY=24 combinations and therefore the formula is n!                

Part 2:

EMMA  EMAM  EAMM  AEMM

AMME  AMEM  MEMA  MAEM

MMEA  MMAE  MAME  MEAM

EMMA   EMMA   EMAM   EMAM   EAMM   EAMM   MMAE   MMAE   MAME   MAME   

 

MEAM   MEAM   MAEM   MAEM   MMEA   MMEA   MEMA   MEMA

 

AMME   AMME   AMEM   AMEM   AEMM   AEMM

The word in bold are the word that are repeated and that is half of the words so because there are two of the same letters you divide the amount of combinations with all the letters different by 2 because there are 2 letters that are the same in the word..

From Emma’s name, a four letter word, we this time only get twelve arrangements. This exactly half of twenty-four,(twenty-four being the total number of arrangements for a four letter word with all the letters different, divided by two because two is the number of letters repeated). I then tried a three letter (TOO) word with two letters repeated

TOO  OTO  OOT

From TOO we get three arrangements, which is the sum of six divided by two (again, six being the number of arrangements for a three letter word with all letters different, and divide by two as it is the number of letters repeated in the word.)

 From this I worked out a formula for the number of arrangements in a word which has two letters the same in it.

This was….

 n!

 2

This formula is demonstrated in the table below.

Part 3:

However, this formula was ineffective when trying to work out the number of arrangements for a word, which has three letters the same in it, and all the rest different.

AAAB   AABA

ABAA   BAAA

The word produced four arrangements, which is 24 divided 6(twenty-four being the number of arrangements of a four letter word with all the letters different, and six being the total number of arrangements of a three letter word and three is the number of letters repeated.)  I worked out that the correct formula for this was…

4!

6 

because if all the A’s were different e.g. A(1) A(2)  A(3)

                                             A(1) A(3)  A(2)

                                                                 A(2) A(1)  A(3)

                                             A(2) A(3)  A(1)

                                             A(3) A(2)  A(1)

                                             A(3) A(1)  A(2)

These would all be different arrangements but seeing as they aren’t they have to be cancel out from 4! so the answer is divided by 6, which is the amount of combinations that are the same.

I found that the arrangements for a word with a double letter were the same as a word with the number of letters of the letter repeated. For example….

I then tried a five letter word with four letters the same.

        AAAAB   AAABA

AABAA   ABAAA

BAAAA

∴ The way to find out the number of arrangements is, for four letters the same, is 5! 

         Because if the A’s were different letters they would form 24 different            24 combinations so the number of combinations for a five-letter word is divided by the amount of combinations the repeated letters make to get the answer.                                                                                                                                

This can then be put into a general formula as……..

                                                        n!

                                                        y!

( ‘y’ equalling the number of repeated letters in the word)

Investigating the number of different arrangements of letters where more than one letter occurs more than once.

For example: words such as XXYY, XXYYXX, and XXXYYY 

The above examples all have more than one letter that repeats more than one time.

Investigation:

Name: “XXYY”

Number of letters: 4

Arrangements:

Total Number of Arrangements: 6

Name: “XXXYYY”

Number of letters: 6

Arrangements:

Total Number of Arrangements: 20

Name: “XXYYY”

Number of letters: 5

Arrangements:

Total Number of Arrangements: 10

Summary of Part 4:

When there is only one letter repeated, the formula obtained was n!/y! (y being the number of repeated letters)

Therefore, for words where more than one letter is repeated, I will try the formula

Examples:

Name: “XXXYYY”

Number of letters: 6

        

Name: “XXYYY”

Number of letters: 5

        

This equation appears to be correct.

• We have therefore arrived at the formula

• The n! is there as the total number of letters, as before. We previously decided that where one letter is repeated a certain number of times, the equation would be

• It is therefore logical to assume that where more than one letter is repeated, n! should be divided by the total number of the first repeated letters multiplied by the total number of the second repeated letters.  So the over all formula is n!/X! x Y! e.c.t.