I then tried to simplify this. By doing so I found that multiplying all the numbers below the number of letters in the given word together, the correct number of arrangements would be calculated. For example, if I was to find the number of arrangements in a seven letter word I would perform the following calculation :
1 x 2 x 3 x 4 x 5 x 6 x 7
This simplified again is 7!.
2. Investigate the number of arrangements of Emma’s name…….
EMMA
EMAM
EAMM
AEMM
AMME
AMEM
MEMA
MAEM
MMEA
MMAE
MAME
MEAM
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, which proves it is correct.
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…….
n!
6
For example, if take the word ABACA, I found that there were twenty arrangements, these were……
ABACA
ACABA
AAABC
AAACB
BCAAA
CBAAA
AABCA
AACBA
ACBAA
ABCAA
ABAAC
ACAAB
BAAAC
CAAAB
BACAA
CABAA
CAABA
BAACA
AACAB
AABAC
This proves my formula. Below is a table to show this.
I soon realised that a pattern was emerging. I found that the arrangements for a word with a double letter was the same as a word with the number of letters of the letter repeated. For example….
A three letter word, with all the letters different would have six arrangements.
A word with three letters the same in it would need to be divided by six to give the correct number of arrangements.
∴ the formula 3! works.
6
I then tried a five letter word with four letters the same.
AAAAB
AAABA
AABAA
ABAAA
BAAAA
∴ The formula for four letters the same would be 4!
24
I the saw another pattern. The denominator in the formula was the same number of arrangements as the number of repeated letters factorilised. For example……..
To find the number of arrangements in a four letter word with three letters the same you would calculate……
4!
3! = Four arrangements.
This can then be put into a general formula as……..
n!
y!
( ‘y’ equalling the number of repeated letters in the word)
To test my formula I will use a six letter word with five letters the same(AAAAAB).
AAAAAB
BAAAAA
ABAAAA
AABAAA
AAABAA
AAAABA
There are six arrangements, which proves my formula is correct.
- A number of X’s and Y’s are written in a row such as……
XX……XXXY……YXXY
Investigate the number of different arrangements of the letters.
I first investigated a number of different words to try and find if a common formula was present, as in the tasks involving Dave’s and Emma’s names.
- A four letter word with three letters the same, the other different.
AAAB
AABA
ABAA
BAAA
There are four arrangements.
- A four letter word, with two different letters the same.
AABB
ABAB
BBAA
BABA
ABBA
BAAB
This gives us six arrangements, which
- A five letter word, with three letters and two letters the same.
AAABB
BBBAA
BABAA
BAABA
BAAAB
ABAAB
ABBAA
ABABA
AABAB
AABBA
There are ten arrangements.
- A six letter word, with two sets of three letters the same.
AAABBB
BBBAAA
ABBBAA
AABBBA
ABABAB
BABABA
BAAABB
BBAAAB
AABBAB
ABBAAB
BBAABA
BAABBA
ABAABB
BABBAA
AABABB
BBABAA
ABBABA
BAABAB
BABAAB
ABABBA
There are twenty arrangements of these letters.
From these words I have discovered a formula that can be used to work out the number of arrangements for a word with more than one type of letter repeated, this formula is found from expanding on the formula used to find the number of arrangements for Emma’s
name (n!) . To find a word with more than one type of letter repeated you must multiply
(y!)
both ‘y’ values together. For example………
To find a the number of arrangements of an eight letter word with two letters and two lots of three letters the same (AAABBBCC) you would calculate
8!____
(3! x 3! x 2!)
I then set about trying to find a formula for this. I quickly came to the decision that all that is needed is letters to represent the number of repeated letters. This gave me the formula:
____n!____
(x! X y! X z!)
(‘x’ represents the number of repeated A’s, ‘y’ represents the number of repeated B’s and ‘z’ represents the number of repeated C’s)
To test my formula I used a five letter word (AABBC), with two lots of letters the same. This gives me the following number of arrangements……….
CABBA
CBAAB
CAABB
CBBAA
CBABA
CABAB
AABBC
AACBB
AABCB
ABABC
ABBCA
ABCBA
ACBBA
ABBAC
ACABB
ACBAB
ABCAB
ABACB
BBAAC
BABAC
BBCAA
BAACB
BACAB
BCAAB
BAABC
BCBAA
BCABA
BACBA
BBACA
BABCA