• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Emmas dilemma.

Extracts from this document...


                                                       Emma’s Dilemma                                            

Arrangements for EMMA:

EMMA        EMAM        EAMM        MMAE

MMEA        MEAM        MEMA        MAME

MAEM        AMME        AMEM        AEMM

Arrangements for MIKE

MIKE                 MIEK                MKEI                MKIE

MEIK                MEKI                IKEM                IKME

IEMK                IEKM                IMEK                IMKE

KEMI                KEIM                KIME                KIEM

KMEI                KMIE                EMIK                EMKI

EKMI                EKIM                EIMK                EIKM

If I choose a word where all of the letters are different there will be more combinations.

There were 24 different possibilities in the arrangement of 4 letters that are all different. That is twice as many as EMMA, which has four letters and 2 the same. I have noticed that with MIKE there were 6 possibilities beginning with each different letter. For instance there are 6 arrangements with MIKE beginning with M, and 6 beginning with I and so on. 6 X 4 (the amount of letters) gives 24, the number 6 may have come from 1 x 2 x 3, the number of letters, and multiplied by four because that is how many numbers there are all together.

The difference between how many combinations of the names EMMA and MIKE is the double M in EMMA. There are 24 possible combinations for MIKE and 12 for EMMA.

I will now investigate further words with all letters different.

I will now test a word with 3 letters:

DOG                OGD                GDO

DGO                ODG                GOD

There are 6 possible combinations, 2 for each letter.

Now I will investigate a 2 letter word.

IT        TI

There are two possible arrangements, one for each letter.

...read more.


MEIK                YMEKI                YIKEM                YIKME

YIEMK                YIEKM                YIMEK                YIMKE

YKEMI                YKEIM                YKIME                YKIEM

YKMEI                YKMIE                YEMIK                YEMKI

YEKMI                YEKIM                YEIMK                YEIKM

This proves my prediction is correct, as there are 24 arrangements that can be made from a four letter word, when this is multiplied by 5 I have the total number of arrangements made from a 5 letter word.

These is the arrangements for MIKE but with a letter ‘Y’ in front of them, therefore the ‘Y’ can move into five different positions throughout the word, therefore that is why we multiply 24 by 5.

I will now investigate more words with one set of repeated letters:


AA, only one combination, this is half as many as a normal 2 lettered word.


EGG                GEG                GGE

There are 3 possible combinations; this is half as many as a normal 3 lettered word, such as DOG. This is because of the double letter.

I will now put this data into a table so it is easier to see al of the information.

Number of letters

Arrangements when all letters are different

Arrangements with one set of repeated letters













From this table I can see that a word with one set of repeats has half as many, or is divided by two, arrangements as a word with the same number of letters and no repeats.

...read more.


        1 x 2 x 3 x 4        or          4!

        1 x 2                      2!

A word with 5 letters and 3 repeated letters:

                                  1 x 2 x 3 x 4 x 5          or     5!

        1 x 2 x 3                             3!        

I now know that a word with:

n letters and 2 repeated letters:



n letters and 3 repeats:



n letters and 4 repeats:



n letters and 5 repeats:



Therefore with n letters and x repeats:

                                   1 x 2 x 3 … n             or     n!

                                    1 x 2 x 3… x                                  x!

Now I will explain what happens when two sets of letters are repeated twice.

                                         AABB - 24 possible combinations

The formula for AABB is 4! / 2! x 2! = 6 possible combinations.

To make it easier instead of using letters as such I will use a’s (any letter) and b’s (any other letter). I will start with AABBB:

        AABBB – 120 combinations

This is because the formula is      5!


n = the number of letters in the word.

a,b,c,d….x = the number of times a single letter is repeated.

So another version of this formula is

(a + b + c + … x)!

Because of this formula I can improve the formula for my original words:

MIKE changes from 4! to          4!


And EMMA changes from        4!         to           4!

                                                   2!                    2!1!1!

Because I must show that there are numbers there, but because 1! equals one it has no effect on the answer to the problem.

I will now finish my investigation with a long word with many repeated letters, MISSISSIPPI:

                                          11!                Because of the number of letters in the word


There are 19958500 combinations

for the word MISSISSIPPI

        -  -

...read more.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Emma's Dilemma

    2 pairs of same letters (AABB) 1 1 2 2 1 3 6 3 4 24 12 6 5 120 60 30 6 720 360 180 The figures in blue are following this formula, showing what will come next, if this formula is used. We know from previous investigations that to find the combinations for all five letters being all different is 5!

  2. Maths GCSE Coursework: Emma's Dilemma

    In the table, I treated the letters as four different letters, like I did for Lucy. E M M A EMMA MEMA MEMA AEMM EMAM MEAM MEAM AEMM EAMM MMYE MMEA AMME EAMM MMEA MMAE AMEM EMMA MAME MAEM AMEM EMAM MAEM MAME AMME There are two M's in the

  1. GCSE Mathematics Coursework - Emma's Dilemma

    I was then able to calculate the numbers of arrangements in names of other lengths, and so extend the amount of results which I had. Table showing the number of different arrangements of letters possible in words of different lengths with no repeated letters Number of letters Number of different

  2. In this piece of coursework I will investigate how many times and ways I ...

    I am now going to investigate the different combinations with 3 letters the same. bbb bbbc bbbcd cbbb cdbbb bcbb cbbbbd bbcb bcdbb bdcbb bbbcd bbbdc bbcdb cbdbb cbbdb dcbbb dbbbc dbcbb dbbcb The different combinations with 3 letters the same No of letters Combinations Fractorial 3 1 3! /3!

  1. Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

    However, with two letters repeated twice, the letters which are repeated can be swapped with their opposites, with-out making a new arrangement. Answer, Two letters repeated Three Times: Six letters: AAABBB BAAABB AABABB BAABAB AABBAB BAABBA AABBBA BABAAB ABAABB BABABA Total: ABABAB BABBAA 20 ABABBA BBAAAB ABBAAB BBAABA ABBABA BBABAA

  2. I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

    I worked out that the total number of arrangements was worked out by multiplying the number of letters by all the numbers under that number. So if I was to work out the number of arrangements for a 4 letter word, I would multiply 4 by 3 by 2 by 1, as they are the numbers below that number.


    1*2*3*4*5 (1*2*3*1*2) =10 For example: 3*2*1 is one and 1*2 is another one. If you calculate the first one, which is 3*2*1=6 and multiply it by 2 you will get 12. I got number 2 from the second one; which 1*2=2.

  2. GCSE Mathematics: Emma's Dilemma

    seeing as "LUCY" & "EMMA" are both 4 letter words; by using the same process I acquired the same amount of results (24). There is however, another method. I could treat both the M's as one collective term. Therefore if you swapped them around, you would get the same word

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work