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Emma's Dilemma

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Emma’s Dilemma

I am going to investigate the number of different arrangements of various groups of letters. I will start with the number of arrangements for words with all the letters being different. For this I will use the name LUCY.

For example, LCUY is one arrangement

YUCL is another.

Here are all the combinations of letters in the name LUCY:

LUCY                 UCYL         CYUL         YCUL

LUYC                 UCYL         CYUL         YCUL

LCUY                 ULCY         CULY         YUCL

LCYU                 ULYC         CUYL         YUCL

LYUC                 UYCL         CLYU         YLCU

LYCU                 UYLC          CLUY         YLUC

There are 24 different combinations of letters with 4 letters being all different. I have noticed that there are six combinations with each letter at the start of the word. So, you do 6x4 to find out how many combinations there are. This means that for all words with different letters, there’s going to be 24 combinations.

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Now I am going to investigate the number of combinations when 2 letters are the same. For this I will use the name EMMA

Here are all the combinations in the name EMMA

EMMA         EMAM          EAMM           MMAE

MMEA         MEAM          MEMA           MAME

MAEM         AMME          AMEM           AEMM

This shows that all 4 letter words with 2 letters the same will have 12 combinations. I noticed that this is half the amount of combinations there was when all the letters were different. I then investigated this to see if it was true with all words with 2 letters the same.


Total number of letters

Total Combinations













As you can see, all of the total number of combinations with 2 letters the same is half the amount of when all the letters are different.

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All of the combinations

AAABB         AABAB         AABBA         ABABA          ABAAB

ABBAA         BBAAA         BAAAB         BABAA          BAABA

There are 10 combinations in total, 6 beginning with A and 4 beginning with B


All of the combinations

AAAAB         AAABA         AABAA


There are 5 combinations in total, 4 beginning with A and 1 with B

Because there are two sets of same letters the previous formula cannot be used. Instead, I have come up with another formula:

Total number of letters factorial/

Number of repeated letters factorial (1) x number of repeated letters factorial (2)

For example

A five letter word like AAABB

This has 3 As and 2 Bs

So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10

A four letter word like aabb

This has 2 As and 2 Bs

So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6

A five letter word like AAAAB

This has 4 As and 1 B

So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5

Five letter words like ABCDE

This has 1 of each letter

So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24

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