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.
I will now draw a table of results:
From this table I can work out the number of combinations for a four letter word, without knowing what it is, I can multiply the previous number of arrangements by the present number of letters in the word, this also works for a 3 letter word and so on. From this theory I predict that a 5 letter word will have 120 arrangements because:
5 Number of letters in word
24 previous number of
arrangements
To prove that this is correct I will use the word MIKEY, for the sake of convenience, but I will only show the arrangements for the letter Y and them multiply my result by 5, the number of letters remaining, because that has worked for all of the other words.
YMIKE YMIEK YMKEI YMKIE
YMEIK 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:
AA, only one combination, this is half as many as a normal 2 lettered word.
EGG:
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.
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. This is shown in my table as a four letter word can be arranged 24 different times, if there is a set of repeated letters then there are 12, it is the same for a 3 and 2 letter word. This can not happen for a single letter word as if 1 is divided by 2 this does not give an integer.
I will now prove why this double letter affects the word:
I have 3 sets of four cards, and four slots to put them in. This can be explained because as there are four letters on each card, only so many letters can be put into each slot. I will show this for the words MIKE, a four letter word with no repeats, EMMA, a four letter word with one set of repeats, and EMMA, a four letter word with one set of repeats but the M’s are distinguishable, therefore it is just like a normal four letter word.
I know that there are 24 choices for MIKE and, using the above technique I have found a formula: 4 x 3 x 2 x 1 = 24
With E M M A there would be 24 choices too because you can distinguish between M and M, i.e. M M is different to M M
But if the cards are E M M A then E M M A and E M M A both become E M M A, E A M M and E A M M both become E A M M, therefore every different pairing will become the same, therefore the total number will half. It will also be half for any word with one repeat.
From this I can predict that if there is one letter repeated 3 times I will divide by 6, because of 1 x 2 x 3 = 6 or 3! = 6, the formula for a 3 lettered word, therefore there are less arrangements for the word.
I shall show how this happens by showing the changing arrangements, depending whether you are able to distinguish between the repeated letters or not. I will use the words BAAA and BAAa. Because I can tell the difference between the A’s, therefore it will have more arrangements. BAAa has 24 arrangements because all of the letters are distinguishable, but if I was to write them out then I would be able to see how many arrangements would be the same if the A’s were all the same.
Arrangements:
BAAa
BAaA
BAAa
BAaA
BaAA
BaAA
If I used the first set of cards then all of these words become BAAA, it is because of this that there will always be six times less arrangements. If there was a word with four repeated letters then I would divide by 24, 1 x 2 x 3 x 4 = 24 or 4!, This is due to most of the arrangements being lost because of the four repeats. It is the same for any number of repeats, if I had five repeats I would be dividing by 120, 1 x 2 x 3 x 4 x 5 = 120 or 5!, so if I had n letters repeated there would be 1 x 2 x 3…x n or n! arrangements.
I already know:
A word with 4 letters and 2 repeated letters:
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!
2!
n letters and 3 repeats:
n!
3!
n letters and 4 repeats:
n!
4!
n letters and 5 repeats:
n!
5!
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!
2!3!
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!
1!1!1!1!
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
2!4!4!1!
There are 19958500 combinations
for the word MISSISSIPPI
