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In this piece of coursework my initial aim is to investigate how many different combinations there are for four letters (e.g. ABCD).

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Jeremy Beales                /

Maths Coursework- Matt’s First Theorem

In this piece of coursework my initial aim is to investigate how many different combinations there are for four letters (e.g. ABCD), I also intend to develop this to investigate the way in which by altering the letters to form other kinds of combinations (e.g. ABCC or AAB) the number is affected. Once I have found the general formulae, I will apply these to harder situations and this is what I am aiming to do. I am trying to find the general formulae which can be applied to all situations we set about this by looking at the simplest scenario first i.e. one letter (e.g.A) moving on to harder problems and by the end I hope to be able to find the possible arrangements for any given word. I will do this by using tables and lists of my results to show the possible combinations and make it easy to compare them and to spot the pattern and try and turn this into a general formula. Once the initial

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One Triple Letter-

3 letters- AAA

4 letters- AAAB




5 letters- AAABC



















No. of letters              3         4         5  

No. of combinations  1         4          20

This gives the formula = n!/3!


If you have a triple letter, then you must divide the n! by 3! because there are 3 repeated letters and so the same will apply as above but in this case there would be three repeats of each combination with triple letters than if there were different letters, e.g. AAAB, if it was ABCD there would be 3! or six times the amount of combinations as there are with ABCD.

This is almost the same formula as above and has only changed in the fact that you have to divide by 3! instead of 2! and this leads me to think that I can easily form a generalised formula, this formula is n!/r! which is true because you need to divide the number of letters by the number of repeated letters because each repeated letter removes 1! options and so three repeated letters removes 3! options .

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Random Selection of Letters without order

In this process I am going to assume that AB is the same BA, this will obviously reduce the number of possible selections, after writing out the possibilities for 5 different letters and 6 different letters I quickly saw that the number was just divided by two, thus giving the general formula (n(n-1)/2

Development of formula-

1st - n! – Where n = number of letters

2nd - n!/ r! – Where r =number of repeated letters

3rd - n!/ r!xs!xt! Where r= 1st group of repeats s= second group and so on

4th - n!/(n-1)! This an be extended to form the formula n!/(n-r)! – Where r = number of letters selected from n

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