I am doing an investigation into words and their number of combinations. I will find formulae and work out the number of combinations for the words.

Introduction

I am doing an investigation into words and their number of combinations. I will find formulae and work out the number of combinations for the words. Originally the task was to find combinations for the word EMMA and several other names, but I decided to look at sequences of letters from the alphabet, which makes it easier to monitor and control. I will start by looking at words with no letters the same, and find a formula for that, beginning with words that are 1 letter long and carrying on to letters that are 5 letters long.

E.g.

ABCD

Next I will look at words that have two letters the same e.g ABBC, using the same method as before of starting with 2 letter words and expanding the amount of letters to 5 or 6 letter words. By doing this I am expanding the investigation and gaining more knowledge of the pattern of formulae.

Throughout the investigation I will use 3 algebraic expressions.

n =number of letters in total within the word.

c= number of combinations found in total

n!=n factorial

Investigation

I am going to investigate the different ways in which letters can be organised and find formulas for all of them.

Letters that are all different

I will now try and find a formula for this pattern. I will put the data in a table to make it clearer first:

I will now prove that 5 different letters has 120 combinations:

ABCDE=24

ABCED=24

ABECD=24

AEBCD=24

EABCD=24

Total=120

I worked out the formula was n! after first discovering the pattern was 1x2; 1x2x3; as so on. From prior knowledge I knew an easier way to write this was n! or n factorial. The factorial pattern is:

1!=0x1=1

2!=0x1x2=2

3!=0x1x2x3=6

4!=0x1x2x3x4=24

Now I will draw a graph to show that I have got the right formula and to see the relationship between the number of combinations and the number of letters even more clearly than in a table. This graph should show strong positive correlation between the two variables and I think they will be directly proportional.

So, yes, the line of results fits the theory and I can say the formulae for words where each letter is different (or 1 letter is the same) is:

2 Letters the same

I will now try and find a formula for this pattern by putting all the data in a table to make it easier to understand and observe:

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 will do a graph to check this is correct and show me the relationship between the number of combinations and the number of letters in the combinations:

I expected less combinations and the graph showed me this. 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. Then I tested this theory against the graph, after first plotting the results I already had. The theory proved to be correct because the line matched the line of results. Therefore I can say the formula for words with 2 letters the same is:

n!/2

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

Now I will do a graph, which will show me the relationship between the number of combinations and the number of letters.

I worked out the formula was n!/2x3 after a while, when I thought of trying n! on the bottom of the formula too, as the ...

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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

Now I will do a graph, which will show me the relationship between the number of combinations and the number of letters.

I worked out the formula was n!/2x3 after a while, when I thought of trying n! on the bottom of the formula too, as the denominator. This would mean that for letters with 2 same letters, the formula can also be written as n!/ 2! or n!/1x2, which makes it clearer to understand the pattern. Therefore, n!/1x2x3 or n!/3! seemed the next step up and it works. Then I checked this using the graph before saying it was the correct formula, and the number of combinations on the graph is proportional to the number of letters, plus the theory of n! fits the graph. 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 combinations

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

Now I will do a graph, which will show me the relationship between the number of combinations and the number of letters.

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, then by comparing it with the graph of results. It did, so I now know the formula for words with 4 letters the same is:

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 ANNA, or MAMA 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

Now I will do a graph, which will show me the relationship between the number of combinations and the number of letters.

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!)

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

Now I’ll draw a graph to illustrate the relationship between the number of combinations and the number of letters in the word:

I could see that n!/a!x2! wasn’t working so 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!)

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, thereby proving the formula is correct:

AABBBBCDE=840

AABBBBCED=840

AABBBBECD=840

AABBBEBCD=840

AABBEBBCD=840

AABEBBBCD=840

AAEBBBBCD=840

AEABBBBCD=840

EAABBBBCD=840

Total=7560

Next, I’ll produce a graph to show the relationship between the number of letters and the number of combinations:

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, thereby proving the formula is correct:

AABBBBBCDE=1512

AABBBBBCED=1512

AABBBBBECD=1512

AABBBBEBCD=1512

AABBBEBBCD=1512

AABBEBBBCD=1512

AABEBBBBCD=1512

AAEBBBBBCD=1512

AEABBBBBCD=1512

EAABBBBBCD=1512

Total=15120

Next, I’ll produce a graph to show the relationship between the number of letters and the number of combinations:

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, but maybe in future I shouldn’t be so hasty because the formula might have been totally wrong. This formula is:

n!/(2!x5!)

Mini 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 y other letters the same is:

N!/(a!xy!)

Conclusion for Investigation

I have now completed the investigation and discovered all the formulae I need to know. If I was following the original task I would now know that:

EMMA has 12 combinations and follows the formula n!/2

LUCY has 24 combinations and follows the formula n!

To improve my investigation I should have done some tree diagrams for each section so I could be sure I didn’t miss any combinations out. The trouble with these is when the numbers get bigger, the size of the diagram expands also and you often run out of space. They can get a bit fiddly and tricky, which is why I used another method instead of:

AABBC=x

AABCB=x

AACBB=x

ACABB=x

CAABB=x

Total=5x

These make sure that you leave no combinations out and also are certain to not take up as much space as it does if you write out each combination or if you do a tree diagram. Trouble with them is you need to know the total number of combinations for the number of letters before (e.g. if you were finding combinations for 5 lettered words you would need to know how many combinations you got for 4 lettered words.) This amount is placed in x and at the end you add all the ‘x’s’ up.

To extend my investigation I could have done many things. I could have continued looking at words, and instead of stopping at words with 2 the same and x others the same, I could have looked at words with 3 the same and x others the same; 4 the same and x others different and so on.

Other thing would have been to look at other ideas for combinations-for example number of flowers in a flowerbed and the number of combinations they can be planted in, or number of different combinations of bingo balls in a bag. The possibilities are endless, but I stopped at 2 letters the same and x others the same because I really didn’t need to know any more-how many words with 3 letters the same and 4 others the same do you come across! Besides, the investigation is not meant to get too big, it should be tidy and compressed to the bare necessities.