Part of The Student Room Group

  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  • Level: GCSE
  • Subject: Maths
  • Word count: 2812
    • View my saved documents

    Emma's Dilemma.

    Extracts from this document...

    Introduction

    Lee Hamilton

    Emma’s Dilemma

    Emma is playing with the arrangements of her name, find out how many combinations there are

    I will write down the combinations of Emma by doing each letter at a time swapping the remainder of the letters around, so when I am further into the experiment I can work out the amount of combinations in word by looking at my result tables.

    4 letters

    3 different

    Arrangements for Emma:

    1) emma

    2) emam

    3) eamm

    4) mmae

    5) mmea

    6) meam

    7) mema

    8) mame

    9) maem

    10) amme

    11) amem

    12) aemm

    There are 12 possibilities when there are 4 total letters and 3 different. From this you could make an hypothesis that the formula could be n (number of letters) x the amount of different letters.

    I used mm as 1 combination instead of making them m1 and m2. This is becuase you are finding different combinations and mm is esacially the same every time and can’t be changed to look different. To explain better you can just think of colour cubes. If you have a green, red, blue and blue and u changed the two blues around it still would look the same.

    You also notice that there are 3 combinations for the letter a and e but 6 for m. This is because it just acts as two letter when it is at the front of the word as the other m can move around in the word.

    ...read more.

    Middle

    What if there was a five-letter word? How many different arrangements would there be for that?

    As I have found that there were 24 arrangements for a 4 letter word with all different letters and that there were 6 ways beginning with one of the letter I predict that there will be 120 arrangements for lucyq, 24 for a, 24 for b, 24 for c; etc. 120 divided by 6 (number of letters) equals 24. Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter. For the sake of convenience I have used lucy and put a q in front of it to show that there are 24 different possibilities with each letter of a 5 lettered name being all different.

    5 letters

    All Different

    Arrangements for lucyq:

    l(ucyq) = 24

    u(lcyq) = 24

    c(luyq) = 24

    y(lucq) = 24

    q(lucy) = 24

    This word has 120 combinations.  I can see that the numbers of possibilities for different arrangements are going to dramatically increase as more different letters are used.

    So as a general formula for names with n number of letters all different I have come up with a formula. With Lucy's name; 1x2x3x4 = 24. With qlucy; 1x2x3x4x5 = 120. However this is expressed as factorial. If I key in (lets say the number of letters all different) factorial 6 it gives me 720, with makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word and it continues to fall in that pattern.

    ...read more.

    Conclusion

    Formula =

    n!/x!y!

    I have come to this formula by looking at my result tables from previous investigations. I worked out earlier that n! Is how to work out how many combinations there are in that certain amount of lettered word and now have found out how to work out words with repeated letters and some more than others. E.g. If there is a 4 of one letter in a word it would be 4! And if there was 3 of another it would be 3!. You then times these two together and they are n! is divided by the sum of them.  

    For example:

    A five letter word like aaabb; this has 3 a's and 2 b's

    So : 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10

    A four letter word like aabb; this has 2 a's and 2 b's

    So : 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6

    A five letter word like aaaab; this has 4 a's and 1 b

    So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5

    Five letter words like abcde; this has 1 of each letter (no letters the same)

    So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24

    All these have been proved in previous arrangements. This shows that my formula works!

    Mississippi can be shown as:

    11x10x9x8x7x6x5x4x3x2x1 / 4x3x2x1(s) x 4x3x2x1(i) x 2x1(p) x 1(m)

    =

    39916800 / 1152 = 34650

    This can be expressed as:

    n!/s! i! p! m!

    ...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

    or


    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. Arrangements for names.

      What if I had aaabb? aaabb aabab aabba ababa abaab abbaa bbaaa baaab babaa baaba There are 10 different arrangements for this sequence. What if I had an arrangement of aaaab? aaaab aaaba aabaa abaaa baaaa There are 5 different arrangements for this sequence. If I go back to aaabb; there are 3 a's and 2 b's in a total of 5 unknowns.

    2. Emma's Dilemma

      of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 ) X etc... Combinations Times repeated X ( Times repeated - 1 ) X etc... For example: To find the number of combinations which can be made from using

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

      of repeated letters and the number of permutations they created: AABB: This word is made up of four letters, with two sets of two repeated letters. This word has 6 different arrangements, as you can only begin with either A or B and then there are three letters left that

    2. EMMA'S DILEMMA

      AAABB 6. ABBAA 2. AABAB 7. BAAAB 3. AABBA 8. BAABA 4. ABAAB 9. BABAA 5. ABABA 10. BBAAA I have found 10 arrangements. The formula for this will be like this, because there aren't any changes to be made because you just have to multiply the repeated letters.

    1. EMMA's Dilemma Emma and Lucy

      The formular is confirmed. What about three pairs of same number The formmular need to be rewritten as a=ni/xixixi There are three xi need to mutiple ni, because there are three pairs of same number. if there are two pair of same number of figures of same number, then there

    2. Emma is playing with arrangements of the letters of her name to see how ...

      RREYK KYRER ERKYR RYKRE RRYKE KYERR ERKRY RYKER RRYEK KRERY EYRRK RERKY RKERY KRREY EYRKR RERYK RKEYR KRRYE EYKRR REYKR RKYRE KREYR EKRRY REYRK RKYER KRYRE EKRYR REKRY RKREY KRYER EKYRR REKYR RKRYE YKERR YERRK YRREK YREKR YKRER YERKR YRRKE YRKRE YKRRE YEKRR YRKER YRERK There are 60 combinations for a five-letter name with one repeated letter.

    1. Emma's Dilemma

      This meant that a five lettered word with four repeats could be expressed as: 5! / 4! = 5 NUMBER OF POSSIBILITIES As this equation worked I came up with generalised equation for my new findings, which is: N! T!

    2. permutations & combinations

      1 2 (2*1) 2 3 (3*2*1) 6 4 (4*3*2*1) 24 5 (5*4*3*2*1) 120 6 (6*5*4*3*2*1) 720 7(7*6*5*4*3*2*1) 5040 8 (8*7*6*5*4*3*2*1) 40320 9 (9*8*7*6*5*4*3*2*1) 362880 The mathematical word 'factorial' helped me calculate that. The word factorial is denoted by the symbol (!). Hence 5!

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