• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Emma’s Dilemma

Extracts from this document...

Introduction

Denise Shaw

Maths Coursework

Emma’s Dilemma

I am using the name Emma playing with the letters in the name to make different arrangements. Emma is a 4 letter word I should get 24 different ways of writing it but because it has 2 letters the same I will only get 12 different ways of writing it.

The ways are:-

1.emma     7.mmae

2.eamm     8.mema

3.emam     9.mame

4.meam    10.aemm

5.maem    11.amme

6.mmea    12.amem

I am now going to use the word Lucy, Lucy also has 4 letters in it but because

...read more.

Middle

3.mto   6.omt

Now I am going to use AA a 2 letter word with 2 letters the same, I will get 1 arrangement.

The 1 arrangement is:-

1.aa

And Jo a 2 letter word with no letters the same gives 2 different ways.

The 2 different ways are:-

1.jo

2.oj

#letters in name

0 letters repeating

1 letter repeating

2

2

1

3

6

3

4

24

12

I predict that if I do a 5 letter word with 2 letters the

...read more.

Conclusion

9.dnyna     19.anydn    29.naynd    39.nnady    49.yannd    59.ynadn

10.dnnya   20.anynd   30.nadny     40.nnday    50.yandn    60.ynnda

I predict for a 6 letter word with no letters the same I would get 720 different arrangements. The way to do this is start from 1 and go all the way up to 6 and times them all together and you get the answer for example 1X2X3X4X5X6 =720 arrangements.

To find out the answer on a calculator the formula is :-

n

shift

x-1

=

This is called factorial

Factorial is the product of all the positive ategers from 1 to a given number for example 4 factorial, usually written 4! Is product 1.2.3.4=24

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Emma's Dilemma

    X 2!) 5 30 = 5! / (2! X 2!) Table 5, Two letters repeated three times: Number of Letters: Number of different combinations: 6 20 = 6! / (3! X 3!) Rule: From these sets of results, I can see that if I expand the rule I found to solve "Two letters repeated twice", then this rule

  2. GCSE maths coursework: Emma's dilemma

    10 arrangement so on ------10 arrangements 112212 121122 112221 122112 Total arrangement= 20 the formula work I expect the arrangements for 8 fig will be a= (1*2*3*4*5*6*7*8)/1*2*3*4*1*2*3*4=70 let's confirm 11112222 11211222 11221212 11121222 11212122 11221221 11122122 11212212 11222112 --------- 15 arrangements 11122212 11212221 11222121 11122221 11221122 11222211 12111222 12122112 12212121

  1. I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

    can be swapped round producing three arrangements starting with either A or B. As you can start with both you therefore can create 6 arrangements as shown below: AABB ABAB ABBA BAAB BBAA BABA This also proves my previous equation incorrect for this word as it would be: 4!

  2. Emma's Dilemma

    Now I will investigate a 5-letter word. LAURA ALURA ARLUA ULARA RLAUA LAUAR ALUAR ARLAU ULAAR RLAAU LARUA ALRUA ARULA ULRAA RLUAA LARAU ALRAU ARUAL UALRA RALUA LAAUR ALAUR ARALU UALAR RALAU LAARU ALARU ARAUL UARLA RAULA LUARA AULRA AALUR UARAL RAUAL LUAAR AULAR AALRU UAALR RAALU LURAA AURLA AAULR

  1. GCSE Maths Project – “Emma’s Dilemma”

    - the same letter appears three times), there should be a formula of Examples: Name: "BBBC" Number of letters: 4 ( Name: "CCCDE" Number of letters: 5 ( * This equation does not appear to work. * The equation for the number of arrangements when one letter occurs twice did however work.

  2. Emma’s Dilemma.

    3 E L N 4 2 3 1 E N L 5 3 1 2 N L E 6 3 2 1 N E L With three letters the pattern of the number of permutations starts to widen out. Altogether there are now 6 different possible combinations for a three-letter name.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work