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Emma's Dilemma

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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.

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