• 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...


Matthew Shaw/10Ds/Mrs Finn      17/07/02    Mathematics GCSE          

 Emma’s Dilemma

On this piece of mathematics coursework called Emma’s Dilema we are going to look at all the different ways that we can arrange names. On this piece of coursework I am trying to achieve to find all the different ways I can arrange names and also I want to find out a good enough formula so that i can find the different arrangements more easily.  

...read more.



3 letters = SAM                                       SIMNO

                 SMA                                       SMNIO

                 ASM                                       SMNOI

                 AMS                                       SIMON

                MAS                                        SMION

                MSA                                        SINMO


Looked at LUCY                                     SONIM



...read more.


 S using this formula

There are 5 letters in SIMONs name so what we do because there is 24 arrangements we times

24 x 5 = 120

So there are 120 arrangements in SIMONs name


Number of Letters

Number of Arrangements













I am going to now investigate names with a repetition of a letter in them for


D,D = 1 , MUM = 3 , EMMA = 12 , KELLY = 60

I have found out that if a name with a letter repetition of 2, then it has half the amount of arrangements, where as a name without a repetition would have more letters.

...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. Peer reviewed

    Emmas Dilema

    3 star(s)

    For example, if I use the combinations ABC and ABCD. ABC has 6 different combinations. If I add a new letter into the combination (D), it means that if I begin with A, there is now 3 spaces for 3 letters to switch around meaning rather than us simply having

  2. Emma's Dilemma

    We can see here that when another pair of the same letter comes into the case, the result is exactly half than what it was before the pair was added (shown in bold RED). Why this works Explaining why this works is difficult, so I will use the help of the previous studies to explain.

  1. Maths GCSE Coursework: Emma's Dilemma

    =24 combinations Letters in brackets can be rearranged I have created a table showing my results: Number of Letters Arrangements Pattern 1 1 x1 2 2 x2 3 6 x3 4 24 x4 4x3x2x1 is known as 4! (four factorial). This is shown on the graph above.

  2. GCSE Mathematics Coursework - Emma's Dilemma

    continued to investigate different formulae which could be used to find the number of arrangements in a word with two sets of repeated letters.

  1. Emma's Dilema

    So: A(BCDE) 24 B(ACDE) 24 C(ABDE) 24 D(ABCE) 24 E(ABCD) 24 120 So it works! A 5 letter word does have 120 arrangements; this is 5�4�3�2�1. This is a factorial, which is expressed as 5!. No. of arrangements factorial answer TIM 6 3�2�1 6 LUCY 24 4�3�2�1 24 ABCDE 120

  2. emmas dilema

    After the first two letters are written there are 2 other letters that can be chosen e.g. if LU has been written then C and Y are left. After three letters have been picked there is only 1 letter left e.g.

  1. Emma's Dilemma

    found a pattern which is known as a factorial, and whilst working this pattern out I found that the pattern of factorials also works this time round. My findings are shown below: 1 2 3 4 5 120 60 20 5 1 1 * 1 * 2 1 * 1


    which is 2.then you will divide 720 by 2, that equals to 360. This can be made simpler like this: 6! / 2! = (1*2*3*4*5*6) / (1*2) = 720 / 2 = 360 If you don't do it like this then the arrangement will be wrong.

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