Now, I’ll prove that there are 120 combinations in a 5 letter word:
ABBCD=24
ABBDC=24
ABDBC=24
ADBBC=24
DABBC=24
Total= 120
I worked out the formula was n!/2 because I put the figures for n! alongside the combination which showed me the combination was exactly half that of the n! so it seemed logical to test n!/2 which worked.
n!/2
Proof
Looking at this the least number of combinations 1 is achieved buy a 2 letter word. This means that the number combination goes 1 3 12 when before it was 1 2 6 24, by removing the 2 all the numbers after are halved and the number of letters same is 2 is this a coincidence? I will find out with 3 letters the same
3 Letters the same
I will now try and find a formula for this pattern. First I will simplify the data by putting it in a table:
I will prove that 6 letter words have 120 combinations:
ABBBCD=20
ABBBDC=20
ABBDBC=20
ABDBBC=20
ADBBBC=20
DABBBC=20
Total=120
This time the least number of combinations is for a 3 letter word which means that the numbers of arrangements for 2 and 3 have been missed out. Looking at the table the original combination has been divided by 6 and 2 x 3 = 6. so in a matter of fact the number of combinations is n!/3x2, 3x2 is really the same as 3! as the last x 1 does not effect the answer so for a bigger number it might be easier to write n!/3! And this is true for 2 letters and no letters the same but until now I haven’t realised it as 2! Is simply 2 x 1 and the x 1 does not effect the answer.
Therefore the formula for words with 3 letters the same is:
n!/3!
Prediction
At this point it is becoming clearer to me the pattern between all the different numbers and formulae. It is a basic statement that all the formulae are involved in n! so in working out the next formulae I will focus on n!. It also seems to me that a denominator divides all the numbers, and the denominator focuses on n! as well. The denominator seems to me to be the factorial of the number of same letters.
E.g.
For each sector (1 letter the same, 2 letters the same, etc) the factorial needs to be divided by something. For example, if you were looking at 2 letters the same and words with 4 letters in them, you would find 4! first, but then you would need to half it (4!=24 and 4!/2!=12). This is because there are half the amount of combinations as there are for the factorial. The same method applies for words with 3 letters the same, there are three times less combinations than the factorial, and so we divided the factorial by 3.
Therefore, I can predict that for words that have 4 letters the same, the formula will be n!/4!.
For example, if I had a 6 letter word with 4 letters the same, I can work out what I think the number of combinations will be:
6!/4!=30
30 arrangements
I will now continue my investigation and investigate the formulae for words with 4 letters the same, words with 5 letters the same, and a few others. We will see if my prediction was right as we go on.
4 Letters the same
I will now try and find a formula for this pattern. First I will simplify the data by putting it in a table:
I will prove that there are 210 combinations for 7 letter words:
AAABCD=30
AAABDC=30
AAADBC=30
AADABC=30
ADAABC=30
DAAABC=30
Total=210
I worked the formula out using my prediction on the previous page, and checked the formula first by seeing if it worked for the results I already had. I now know the formula for words with 4 letters the same is:
n!/4!
This is because now the least number of combinations is achieved by a 4 letter word so the arrangements for 1 2 3 and 4 have been missed out so the formula is n!/ 4 x 3x 2 x 1 which is the same as n! / 4!
Mini Conclusion
As a mini conclusion for this part of the investigation, I will respond to my prediction and say whether I got it correct and summarise.
In my prediction, I said that there was a pattern between all the equations I had discovered at that point. I followed on by showing the patterns and explaining what I thought the pattern was. At that point I did not give a formula because I felt it would be better to see if the rest of the formulae followed that pattern and then summarise, as I am doing here by giving a formula.
In the prediction, I said that I thought the pattern was n factorial divided by the factorial of the number of same letters in that section of the investigation. I predicted that for words with 4 letters the same the formula would be n!/4! and worked out that for words with 6 letters in that had 4 letters the same, the number of combinations found would be 30. I will now draw a table to see if my prediction was right:
Therefore I can see my prediction was correct. The formula for letters with a certain numbers the same is:
n!/a!
(Where a!= the number of letters the same)
2 lots of 2 Letters the same
I have enough information I need for words with a number of letters the same but then I realised this doesn’t give me a formula for certain words, such as the name MATT, where there are two lots of 2 letters that are the same, so I thought to extend my investigation and make it more fruitful, I will include a glance into these kinds of words and work out a formula for them.
I will now try and find a formula for this pattern. First I will simplify the data by putting it in a table:
Now I shall prove that 7 different letters in a word with 2 lots of 2 letter the same has 1260 combinations:
AABBCDE=180
AABBCED=180
AABBECD=180
AABEBCD=180
AAEBBCD=180
AEABBCD=180
EAABBCD=180
Total=1260
The pattern for these types of words seems to also follow n factorial, which I was expecting because they are not that different from the words with a certain amount of letters the same. However, the pattern is different and doesn’t follow n!/a!. I thought that if I focused on n! and took the fact that it was the numerator for granted, I might find a formula more quickly than if I just tried to find any formula that worked. After a while I found that if I did the basic formula-n!/a! (Presuming a is the number of letters the same) I got twice the amount I had for the combination. From there it was quite simple-I tried doing (n!/a!)/2 and got the right amount for the combinations. Then I worked out that n!/(a!x2!) gave me the same result but was a bit simpler to follow. Therefore it seems clear that the formula for 2 lots of 2 letters is:
n!/(a!x2)
Proof
I no from previous experimenting that for 1 pair of letters the formula is n!/2! so I used this to try and expand on 2 pairs and I discovered that it is (n!/2!)2 which is that same as n!/(2! x 2). This is because the number for the second pair has reduced the number of combinations by 2 (which is possibly 2! But do not no yet) just like the first one did.
At this point I would like to note something that crossed my mind as I worked out the formula. I realised that the (a!x2!) part was the same as (2!x2!) in this formula which is the same as what I am investigating-2 lots of 2 letters the same. It would be worth seeing if in my next sections whether this pattern continues.
2 letters the same and 3 letters the same:
Now I will look at words with 2 of the same letter the same and 3 of another letter the same. (eg AABBB)
Now I’ll try to find a formula for this pattern, bearing in mind my thoughts from the last section. First, to make things clearer, I’ll simplify the data by putting it in a table:
I will now check to see whether I can prove this is the right formula and 3360 is the right number of combinations for 8 letter words:
AABBBCDE=420
AABBBCED=420
AABBBECD=420
AABBEBCD=420
AABEBBCD=420
AAEBBBCD=420
AEABBBCD=420
EAABBBCD=420
Total=3360
I decided to give my theory on the last section a go and try n!/(a!x3!) (where a!=2 and 3! is from 2 the same and 3 the same) This worked and therefore the formula for words with 2 letters the same and 3 letters of another letter the same is:
n!/(2!x3!)
This formula works because now on top of the formula there is a divide by 6 and I no from before this is the same as 3! as the combinations for 3 and 2 are missing so the whole number is smaller by a factor of 3 x 2 which is the same as 3!.
2 letters the same and 4 letters the same
Now I’ll simplify the data by putting it in a table, therefore allowing me to see it more clearly:
I will now prove that 9 lettered words have a combination of 7560
AABBBBCDE=840
AABBBBCED=840
AABBBBECD=840
AABBBEBCD=840
AABBEBBCD=840
AABEBBBCD=840
AAEBBBBCD=840
AEABBBBCD=840
EAABBBBCD=840
Total=7560
It wasn’t that hard to work out the formula for this one, since I am starting to get a good idea of the pattern. Next, I will write up a prediction for words with 2 the same and others with a certain amount the same.
For this section, I can say I have proved and tested the formula and found that it works. This formula is:
n!/(2!x4!)
Prediction
I shall now predict what I think the pattern is between the sections I have just done that follow 2 letters the same and x other letters the same. I already know all the formulae in the whole investigation is going to follow n!/a! and now I think I have found a successful method for letters with 2 pairs the same; letters with a couple the same and 3 others the same; and letters with 2 the same and 4 others the same. It seems that after n! you take the number of letters the same (eg 2,2; 2,3; 2,4;) and multiply them together, all in their factorial form. Then you divide n! by the total.
Therefore I can predict that for words with 2 letters the same and 5 others the same, the formula would be n!/(2!x5!).
I will now prove this and conclude this part of the investigation with a mini conclusion. We shall then see if my prediction is correct.
2 letters the same and 5 letters the same
Now I’ll simplify the data by putting it in a table, therefore allowing me to see it more clearly:
I will now prove that 10 lettered words have a combination of 15120:
AABBBBBCDE=1512
AABBBBBCED=1512
AABBBBBECD=1512
AABBBBEBCD=1512
AABBBEBBCD=1512
AABBEBBBCD=1512
AABEBBBBCD=1512
AAEBBBBBCD=1512
AEABBBBBCD=1512
EAABBBBBCD=1512
Total=15120
In my prediction I predicted that the formula would be n!/(2!x5!) and after testing and trying this formula I can now say it is. I have proven this as well. I worked the formula out by testing out the predicted formula straightaway without looking at any other pattern. The formula was right, this formula is:
n!/(2!x5!)
This formula works because now the number is smaller by a factor of 5x4x3x2x1 which is 5! As the smallest number of combinations is achieved by 5 letters so the rest (1, 2, 3, 4 letters) are not needed in this so they are removed.
Conclusion
I will now refer back to my prediction and say whether the formula was correct and summarise.
In my prediction I said that I thought the formula continued to have n!/a! in it but the denominator had the 2 numbers of letters the same multiplied. (e.g. for 2 letters and 2 other letters the same the 2 and 2 are multiplied.)
This proved to be correct. I also said that the formula for words with 2 letters the same and 5 other letters the same the formula would be n!/(2!x5!), which it was.
I can now say the formula for words with a letters the same and b other letters the same is:
N!/(a!b!)
Investigating the arrangements of letters in words
Avninder Gidar