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  • Level: GCSE
  • Subject: Maths
  • Word count: 4266

I have been given a problem entitled 'Emma's Dilemma' and I was given the following information: 'Emma and Lucy are playing with arrangements of their names

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Emma’s Dilemma

I have been given a problem entitled ‘Emma’s Dilemma’ and I was given the following information:

‘Emma and Lucy are playing with arrangements of their names. One arrangement of Lucy is:

L        U        C        Y

A different arrangement is:

Y        L        C        U

Part 1:

Investigate the number of different arrangements of the letters of Lucy’s name.

Part 2:

Investigate the number of different arrangements of the letters of Emma’s name.

Part 3:

Investigate the number of different arrangements of various groups of letters.’

So basically I had to investigate how the number of permutations for words alters depending on the letters that make up that word, starting off with looking at Lucy and then comparing it to Emma, and after that extending it to look at any word with various letters. I aim to produce a formula to produce the number of permutations.

Part 1- Investigating Lucy:

The name Lucy comprises of four different letters that we can quite simply rearrange to produce a number of arrangements. Firstly I worked out the number of arrangements that I could get from the letters of ‘Lucy’:

LUCY                LUYC                LYUC                LYCU                LCUY                LCYU                ULCY                ULYC                UCLY                UCYL                UYLC                UYCL

CLUY                CLYU                CYLU                CYUL                CUYL                YLUC                YLCU                YUCL                YULC                YCUL                YCLU

I found that the name Lucy can be arranged 24 different times as shown above. This would work for any four letter word as long as all the letters were different. So after realising this, I decided to investigate this further. I then looked at words with different numbers of letters, with all the letters making up the word being different. I came up with the following:

One Letter:

For a word with one letter, there is only 1 combination as shown below:


Two Different Letters:

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Arrangements with a Repeated Letter:






After looking at this table, I could see that there was a pattern. The arrangements of words with different letters had exactly double the arrangements of words with one repeated letter, both words having the same amount of letters.

So when two words have the same amount of letters, one having totally different letters and one having one repeated letter, the one with the different letters will have twice the number of possible arrangements of the one with a repeated letter.

So the formula calculating the number of arrangements (Y) for a word, such as Emma with X letters but X-1 different letters is:

Y                =        X!         /         2

So we can look in general at how having letters the same affects the number of arrangements. So far we have looked at ABCD and ABBC.


When all the letters are different, the amount of permutations is not restricted and you are free to move the letters into any position.


When two of the letters are the same, the number of restrictions is reduced by half. When listing the arrangements you, you can’t complete half of the ones you would do if the two letters were different as the letters have already been in that position, even if it may have been the first B in the word for example and not the second B, the arrangement can’t be completed as the two Bs are not distinguished between.

As it seemed interesting that the number letters that were repeated in the word and the number you had to divide by in the formula were both 2, I thought it might be interesting to investigate if this formula further changed when I changed the number of repeated letters. I decided to investigate the number of arrangements for words with three letters repeated:  


This word is made of three letters, all of which are the same. Therefore there can be no possible arrangements other than 1:



This word is made up of four letters, three of which are repeated. This should greatly restrict the number of arrangements as there is only one letter that can move round as the rest are the same. Below are the 4 arrangements that could be produced:

ABBB                BABB                BBAB                BBBA


The word ‘ABCCC’ is a five letter word comprised of three letters that are the same. The following 20 permutations could be produce using this word:

ABCCC        ACBCC        ACCBC        ACCCB        BACCC        BCACC        BCCAC        BCCCA        CABCC        CACBC        CACCB        CBACC

CBCAC        CBCCA        CCABC        CCBAC        CCACB        CCBCA        CCCAB        CCCBA

So if we put this into a table in order that the data is easier to view:

Table 4        

Letters in the word with 3 repeated letters








So we can now look at this data in relation to the formula I produced earlier for words with two repeated letters. The formula stated that the number of combinations (y) would be equal to letters in the word (x) factorial, divided by two (y=x!/2). So if we look at the equation and if it would be correct in a word where there are three of the same letters.

Table 5

Letters in the word including three of the same letters (x):




Answer to the formula for combinations (y=x!/2):




The Correct Answer:




The formula for words with two letters the same quite clearly doesn’t work for words with three letters the same. This seemed interesting as the formula obviously changed according to the number of repeated letters. I now wanted to find out how the formula changed according to number of the same letters.

So, considering that in the formula for words with two letters the same, you had to divide the factorial by two, I thought that there may be a link between the number of times the letter was repeated and the number you had to divide by in the formula. So I then tried replacing the 2 in the previous formula with 3 as there was a letter that came up in the words 3 times. This gave me the formula:

Y        =        X!        /        3

I then used this equation and worked out if it was correct or not, when looking at words with a letter that occurred three times:

Table 6

Number of Letters (x):




Answer to Y = X! / 3:




Correct Answer:




This too was not the answer, but did bring up a clear pattern. The answer to the formula ‘Y = X! / 3’, is exactly double the correct answer on all three examples I have used. Therefore if we look at the formula instead of dividing the factorial of x by three, we can divide it by two and get the correct answer as shown below:

Table 7

Number of Letters (x):




Y = X! / 6:




The Correct Answer:




...read more.


AABBC        AABCB        AACBB        ABBCA        ABCBA        ABBAC        ABCAB        ABACB        ABABC        ACABB        ACBAB        ACBBA        BAABC        BABAC        BABCA        BAACB        BACAB        BACBA        BBCAA        BBACA        BBAAC        BCAAB

BCABA        BCBAA        CAABB        CABAB        CABBA        CBABA        CBAAB        CBBAA

So looking at the word I can then apply the same rules for this word. The word is made up of 5 letters so we

The maximum value is 5! which is equal to 120. Then we look at the letters that make up the word. One letter occurs only once so we get the value of 1! which is equal to 1. There are also two letters that occur two times and so we then get the values of 2! and 2! which are both equal to 2. We then multiply the 3 factorial together:

( 1 ) x ( 1 x 2) x ( 1 x 2 )                = 1 x 2 x 2        =  4

So we then put the two values together and workout the answer for the number of permutations for AABBC as follows:

Y        =        5!        /        2!        x        2!        x        1!

Y        =        120        /        4

Y        =        30

This proves that the formula that I came up with is correct for working out the number of different arrangements for any word no matter what letters it contains.

The final general formula is shown below:

Y        =        X!        /        A!        x        B! _        x _        C! Etc.


Y is the number different permutations.

X is the number of letters in the word.

A is the number of repetitions of any letter.

B is the number of repetitions of any second letter.

C, D, E etc. is the number of repetitions of the remaining letters.

So applying this formula I can work out the number of different permutations for any word as shown below:


Y        =        X!        /        A!        x        B!         x         C! Etc

Y        =        9!        /        4!        x        3!         x         2!

Y        =        362880        /        24         x        6        x        2

Y        =        362880        /        288

Y        =        1260

There are 1260 different permutations for the word AAAABBBCC.


Y        =        X!        /        A!        x        B!         x         C!        x        D!

Y        =        16!        /        3!        x        4!         x         5!        x        4!

Y        =        1307674368000        /        6        x        24        x        120        x        24

Y        =        1307674368000        /        414720

Y        =        3153150

There are 3153150 different permutations for the word AAABBBBBCCCCCDDDD

Jacob Large 10PD Mr Neary Mathematics Emma’s Dilemma Coursework

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