# 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.

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Introduction

## 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

1 letter | 2 letters | 3 letters | 4 letters |

A | AB | ABC | ABCD=6 |

BA | ACB | ABDC=6 | |

Total=1 | ADBC=6 | ||

Total=2 | CBA | DABC=6 | |

CAB | |||

Total= 24 | |||

BAC | |||

BCA | |||

Total=6 |

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

No. of letters | Combination (C) | Working |

1 different letters | 1 | 0x1=1 |

2 different letters | 2 | 1x2=2 |

3 different letters | 6 | 1x2x3=6 |

4 different letters | 24 | 1x2x3x4=24 |

n different letters | n! | |

I believe the formula is n! I will now test this and predict the combination for 5 different letters. | ||

5 different letters | 120 | 1x2x3x4x5=120 or 5! =120 |

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

ABCDE=24

ABCED=24

ABECD=24

AEBCD=24

EABCD=24

Middle

Total=1

AABAA

AAACAB=5

ABAAA

AACAAB=5

BAAAA

ACAAAB=5

CAAAAB=5

Total=5

Total=30

I will now try and find a formula for this pattern. First I will simplify the data by putting it in a table:

## Number of letters | Combination (C) | n! | Nth term | |

4 letters | 1 | 24 | 3! /1x2x3=1 | |

5 letters | 5 | 120 | 4! /1x2x3=4 | |

6 letters | 30 | 720 | 5! /1x2x3=20 | |

n letters | n | n!4! | ||

I believe the formula is n!/ 4!. I will now test this and predict the combination for 7 letters. | ||||

7 different letters | 210 | 5040 | 7!/1x2x3x4=210 |

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:

Number letters same | Actual formula | Predicted formula |

1 | n!/1! | n!/1! |

2 | n!/2! | n!/2! |

3 | n!/3! | n!/3! |

4 | n!/4! | n!/4! |

5 | n!/5! | n!/5! |

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.

4 letters | 5 letters | 6 letters |

AABB | AABBC=6 | AABBCD=30 |

ABBA | AABCB=6 | AABBDC=30 |

BBAA | AACBB=6 | AABDBC=30 |

ACABB=6 | AADBBC=30 | |

BAAB | CAABB=6 | ADABBC=30 |

ABAB | DAABBC=30 | |

BABA | Total=30 | |

Total= 180 | ||

Total=6 |

I will now try and find a formula for this pattern. First I will simplify the data by putting it in a table:

Number of letters | Combination (C) | n! | n!/2 | Working |

4 letters | 6 | 24 | 12 | 4!/(2!x2!)=6 |

5 letters | 30 | 120 | 60 | 5!/(2!x2!)=30 |

6 letters | 180 | 720 | 360 | 6!/(2!x2!)=180 |

N letters | n | n! | 2n | n!/(2!x2!) |

I think the formula is n!/(2!x2!) I will now test and predict the combination for 7 letters: | ||||

7 letters | 1260 | 5040 | 2520 | 7!/(2!x2!) |

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)

5 letters | 6 letters | 7 letters |

AABBB | AABBBC=10 | AABBBCD=60 |

ABABB | AABBCB=10 | AABBBDC=60 |

ABBBA | AABCBB=10 | AABBDBC=60 |

ABBAB | AACBBB=10 | AABDBBC=60 |

BAABB | ACABBB=10 | AADBBBC=60 |

BABAB | CAABBB=10 | ADABBBC=60 |

BABBA | DAABBBC=60 | |

BBAAB | Total=60 | |

BBBAA | Total= 420 | |

BBABA | ||

Total=10 |

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:

Number of letters | C | n!/a!x2! | n!/a!x3! | Working | ||

5 letters | 10 | 30 | 10 | 5!/(2!x3!)=10 | ||

6 letters | 60 | 180 | 60 | 6!/(2!x3!)=60 | ||

7 letters | 420 | 1260 | 420 | 7!/(2!x3!)=420 | ||

n letters | n | n!/a!x3! | ||||

I think the formula is n!/(2!x3!). I will test and predict the combination for 8 letter words. | ||||||

8 letters | 3360 | 10080 | 3360 | 8!/(2!x3!)=3360 |

Conclusion

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.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

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