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• Level: GCSE
• Subject: Maths
• Word count: 2746

# Emma&amp;#146;s Dilemma

Extracts from this document...

Introduction

Emma’s Dilemma

I have been given the task of finding different combinations for different names with different amount of letters. I am to find the right formula that will work for all names.

I have been given two names. “Lucy” and “Emma” and I am to find a formula that will hopefully help me to work out all the combinations for any word that is given to me.

Firstly I will take the name Lucy because I have notice that in the name Emma there is a repeated letter and so I will leave this for later.

Lucy is a four letter word and it would make sense to firstly start with a two letter word and work my way up to four so that I may be able see a trend or pattern.

I will take the name “Jo”: -        jo

Oj

As you can see there are only 2 different combinations for the name Jo.

The next name I will take is “Sam”: -        sam        ams        msa

Sma        asm        mas

The name Sam has 6 different combinations.

Now I have come to a four letter word I can work out the combinations for “Lucy”: -        lucy                ulcy                cylu                yluc

Luyc                ulyc                cyul                ylcu

lcuy                uylc                cluy                ycul

lcyu                uycl                clyu                yclu

lycu                ucyl                cuyl                yucl

lyuc                ucly                culy                yulc

As you can see the name Lucy has 24 different combinations.

Middle

Now I can use “Emma” and see how many different combinations there are for this name: -        emma                aemm                mmae

eamm                amme                mmea                                                                emam                amem                meam                                                                                                mema                                                                                                mame                                                                                                maem

The name “Emma” has 12 different combinations. I can now put these results into another table and then see if I can recognise a pattern that will help me to find a formula.

A pattern can’t be recognised straight away so I can see if Factorial Notation will work again.

4! = 1 x 2 x 3 x 4 = 24

This as you can see does not work but what I have seen is that if i then divide the answer of 24 by 2 then i will get the answer that I am looking for.

24 / 2 = 12

I will now test with 2 and 3

2! / 2                                        3! / 2

= 1 x 2 / 2                                = 1 x 2 x 3 / 2

= 2 / 2                                = 6 / 2

= 1                                         = 3

This does work for all of these and will probably work for the rest. The formula is: -

“N! / 2”

I will now do this for the 5-letter word with two letters the same.

The word will be “bemma”. I will first use my formula to find it out and then I will test it by finding out all the different combinations.

N! / 2

= 5! / 2

= 1 x 2 x 3 x 4 x 5 / 2

= 120 / 2

= 60

I believe that when I work out all the combinations there will be 60 different combinations.

Let me see: -        bemma        ebmma        mbema        abemm                                        bemam        ebmam

Conclusion

N!/X!

= 5!/3!

= 1 x 2 x 3 x 4 x 5 / 1 x 2 x 3

= 120 / 6

= 20

Again we get the right answer which is good news it works for all of these, so we now do not need the other three formulas that I made up earlier.

There is one alternative that I have discovered with this term.

If you have a word with two pairs of 2 letters the same then you will find that “N!/X!” does not work. But I have found a way round this. All you need to do is add a “Y!” this will make it work.

E.g. If we take the word “aabb” then we will try the formula that I have just discovered: -

N!/X!

= 4!/2! ?

I can see that there are two different letters repeated and I can’t substitute these into the equation. All I need to do is make the equation “N!/X!Y!

So I will try this and see if it works: -

N!/X!Y!

= 4!/2!*2!

= 1 x 2 x 3 x 4 / 1 x 2 * 1 x 2

= 24 / 2 * 2

= 24 / 4

= 6

This formula works with this as there are only 6 different combinations for this word. They are: -        aabb                baab                                                                                 abab                baba

abba                bbaa

as you can see there are only 6 different combinations.

If you are given a word such as “aabbcc” then you would simply add another letter to the term. It would become “N!/X!Y!Z!”.

I have tried to explain this in the simplest way possible and hope that it is understandable. My final formula is “N!/X!” and more letters can be added if there is more than one letter repeated.

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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