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

Emma's Dillemma - Rearranging Emma's Name in different permutations

Extracts from this document...

Introduction

Arrangements for Emma: emma emam eamm mmae mmea meam mema mame maem amme amem aemm There are 12 possibilities; note that there are 4 total letters and 3 different. What if they were all different like Lucy? Arrangements for Lucy: lucy ucyl cylu ycul luyc ucly cyul yclu lcuy ulcy culy yulc lcyu ulyc cuyl yucl lyuc uycl clyu ylcu lycu uylc cluy yluc There are 24 different possibilities in this arrangement of 4 letters that are all different. That's twice as many as EMMA, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For instance there are 6 arrangements with Lucy beginning with L, and 6 beginning with u and so on. 6 X 4 (the amount of letters) gives 24. What if there were 4 letters with 2 different? Arrangements for anna: anna anan aann nana naan nnaa There are 6 arrangements for aabb. From 2 different letters to all different letters in a 4-letter word I have found a pattern of 6, 12 and 24. ...read more.

Middle

So as a general formula for names with x number of letters all different I have come up with a formula. With Lucy's name; 1x2x3x4 = 24. With qlucy; 1x2x3x4x5 = 120. This is called a factorial. There is a button on most scientific calculators which basically does this formula for you according to the number of letters in the word. If I key in (assuming all the letters are different) factorial 6, I get it gives me 720, which makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word, and it continues to follow that pattern. Total Letters (all different) Number of Arrangements 1-1 2-2 3-6 4-24 5-120 6-720 So now that I've explained the pattern of general x lettered words, what do I do if any letters are repeated? Like in Emma; it has 4 letters but 2 of which are the same. 4 factorial equals 24, but I could only find 12, which means that there must be more to it that just factorial in that way. ...read more.

Conclusion

I came up with; The total number of letters factorial, divided by the number of a's, b's ect factorised and multiplied Formula for emma= 4!/1!x1!x2!=48 /4=12 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. Total letters (all different!) Number of A's Number of B's Number of arrangements 1 1 0 1 2 - - 2 3 - - 6 4 - - 24 5 - - 120 3 2 1 3 4 2 2 6 5 2 3 10 6 2 4 15 3 3 0 1 4 3 1 4 5 3 2 10 6 3 3 20 ...read more.

The above preview is unformatted text

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 dillemma

    For a two letter word with all the letters different there are 2 permutations. For a three letter word with all the letters different there are 6 permutations. And for a four letter word with all the letters different there are 24 permutations.

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

    multiplied by the previous number c (number of combinations) equals the next c. Eg. C 2 6 24 x 2 3 4 4 ? 6 ? 24 I can write this as nfactorial divided by xfactorial (x standing for the number of arrangements for 1x and ny). Or n? ??

  1. I will find all the different combinations of Emma's name by rearranging the letters. ...

    MMM MMMA MMMAB MMM MMMA ??? MAMM AMMM MMAM Here are my results: Name # Of Letters # Of Arrangements MMM 3 1 MMMA 4 4 In comparison with my original results, the numbers of arrangements are divided by 6.

  2. Investigating the arrangements of letters in words.

    AABBCBB=15 AABBBDBC=105 ABBBAB AABCBBB=15 AABBDBBC=105 ABBABB AACBBBB=15 AABDBBBC=105 ABABBB ACABBBB=15 AADBBBBC=105 BABABB CAABBBB=15 ADABBBBC=105 BBBAAB DAABBBBC=105 BABBAB Total=105 BABBBA Total=840 BBABAB BBABBA BBBABA BBAABB BAABBB Total=15 Now I'll simplify the data by putting it in a table, therefore allowing me to see it more clearly: Number of letters C n!/2!

  1. Dave's Dillemma.

    This gives us a formula: x!/n! x is the total number of letters in the name chosen, whereas n is the number of letters which are the same. I am going to test the formula on the name EMMA: 4!

  2. Investigate the number of different arrangements of the letters of Lucy's name.

    If Y is the last letter, it has only one possibility where it can be placed. This does not necessarily have to happen in this order, it can happen in any order, but it has to follow the same theory.

  1. Find out how many different arrangements of letters there are in a name or ...

    I think I have found the equation. ?! = (no. of letters in factorial form) ?! = (no. of letters repeating in factorial form) Or to look at it in an easier way... NAADUE: 720 � 2 = 360 CAAAA: 120 � 24 = 5 EMMA: 24 � 2 =

  2. Investigate the number of arrangements of Dave's name.........

    I then tried to simplify this. By doing so I found that multiplying all the numbers below the number of letters in the given word together, the correct number of arrangements would be calculated. For example, if I was to find the number of arrangements in a seven letter word I would perform the following calculation

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