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

Emma's Dilemma

Extracts from this document...


Emma’s Dilemma

        Emma and Lucy are experimenting with the arrangements pf the letters in their names.

        I am going to look into how many arrangements of the word Lucy there are; I am then going to find to find all of the arrangements for words with other numbers of letters in them.

        I am looking for a link between the number of letter in a word and the number of different arrangements of the letters there are.


I am starting off with the word Lucy; I am going to write out all the different arrangements of the letters in the name.

 LUCY                LUYC                LYUC                LYCU                LCUY                LCYU        

 ULCY                ULYC                UYLC                UYCL                UCLY                UCYL        

 CYLU                CYUL                CULY                CUYL                CLYU                CLUY        

 YLUC                YLCU                YCUL                YCLU                YULC                YUCL

        There are 24 different combinations for the word Lucy.

I can see from this that the first letter in any arrangement of the letters can be one of N letters (where N is the number of letters in the word). Therefore the second letter can be one

...read more.


        I predict that words with 1 letter repeated twice will have fewer arrangements than words with no repeated letters. I think this because for every combination of letters there will be another that is exactly the same because all you have done is swap around the two repeated letters.


EMMA        EMAM        EAMM        MMAE        MAME        MEAM

MAEN                MEMA        MMEA        AEMM        AMEM        AMME EMMA        EMAM        EAMM        MMAE        MAME        MEAM

MAEN                MEMA        MMEA        AEMM        AMEM        AMME

The letters in italics         have already been written out but using the other m, so therefore there is half the amount of possible combinations. This means that to get the number of possible combinations of a word with one set of letters repeated twice you would have to divide by 2 so that you get half the number of results.

Therefore the expression used  to work out the number of possible combinations of a word with one set of letters repeated twice is:

N!This means that you would divide the factorial of

  1. the length of the word by 2.

I am going to test

...read more.


           6!    _          _        = 90

             2!  2!  2!

I am not going to write out all of the combinations for Hannah because I do not feel that I need to.


        I am now going to test my theory on the name PHILLIP because it has a different number of letters to the word HANNAH.

           7!    _          _        = 630

             2!  2!  2!

Both of these calculations seem to work because they come up with reasonable answers when you compare them to the original N! of the word.

I can see that both of these words work with the expression:


R1! R2!

This expression means that you divide the factorial of the number of letters (N!) by the factorial of the number of first set of repeated letters (R1!) by the factorial of the second set of repeated letters (R2!), this can be done with however many sets of repeated letters there are.


I am lastly going to try my expression with the name ANASTASIA

Because it has a lot of letters in the word and two sets of repeated letters:

           9!    _          _        = 7560

               4!  2!

So the expressions for working out the number of combinations available for any word are these:

N!                -for a word without any repeats

N!-for a word with one set of repeated letters


      N!         _        -for a word with any amount of  sets of repeated letters

R1! R2!

...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

    I know my rule works for Four letters and below, and trying to write out the combinations of any number higher than this ( even 5 letters which would produce 120 different combinations ) would only waste time, and effort.

  2. Emma’s Dilemma

    Step 3. We then have the formula of Number of letters! (L!) / number of same letters! (SL!) = A We can demonstrate this again with the five-letter word of ABBBB, with four letters the same L! / SL!

  1. Emma's Dilemma

    So the formula to find the total arrangement of a word which has a letter repeated four times is n!/24. I have found the formulas for finding the total arrangement of words, which have a certain number of repeated

  2. Emma's Dilemma.

    Therefore as the name RALPH has 120 arrangements so as DAVID has 2 letters the same a shall be half of RALPH. Therefore as Ralph has 120 arrangements the name DAVID will have 60 arrangements. 120/2=60 ARRANGEMENTS FOR DAVID 1)DAVID 2)DAVDI 3)DADVI 4)DADIV 5)DAIVD 6)DAIDV 7)DDVIA 8)DDVAI 9)DDAIV 10)DDAVI 11)DDIVA

  1. Emma's dilemma.

    This has three combinations Four letters MEEE EMEE EMEE EEEM This has four combinations Number of combinations Number of letters in word 1 1 2 2 3 3 4 4 I have observed that in all words that only have one different letter than the rest, the number of letters

  2. Emma's Dilemma

    So, by using factorial notation (!) I can predict the number of different arrangements for words where all the letters are different. 6-letter word: 6 ? 5 ? 4 ? 3 ? 2 ?1 = 720 different arrangements. 7-letter word: 7 ? 6 ? 5 ?

  1. Emma's Dilemma

    name, one could simply have multiplied 4 by 3 by 2 by 1 (or, as it is more commonly known, "4!") instead of listing every single possibility of the letters L, U, C and Y as I did. This applies for fewer or more letter names, words or combinations.

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

    of letters - 1 ) X ( No. of letters - 2 ) X etc... combinations ( times a letter X ( times a different letter has been repeated ) has been repeated ) From this rule, all that is needed to be done, is to introduce a small mathematical function, called "FACTORIALS ( !

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