• 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...


Maths GCSE Coursework.

Emma’s Dilemma

Emma is playing with the arrangements of the letters of her name.  The standard arrangement is ‘EMMA,’ another arrangement is ‘MEAM,’ and a different arrangement is ‘EAMM.’

  1. Investigate the number of different arrangements of the letters of EMMA’s name.

Emma has a friend called LUCY

  1. Investigate the number of different arrangements of the letters of LUCY’s name.

Choose some different names

  1. Investigate the number of different arrangements of the letters of the names you have chosen.

A number of X’s and Y’s  in a row such as : XXX……..XXYY……..YYY

  1. Investigate the number of different arrangements of the letters.

Shown below is the full list of the combinations of her name:













As you can see there are 12 combinations of Emma’s name.  This process of writing all of the combinations out is very slow and very tedious; therefore it would be a lot better if we had a formula to work out the number of combinations.  This would prove especially useful if you had a name like Bartholomew, which would take you ages to work out the number of combinations.

...read more.






















There is however the problem of repeating letters, like the two m’s in EMMA.  This make the number of possible formulas a lot less, especially in longer names.  However an example for 3 letters is shown below:







= 6 Combinations




= 3 Combinations

As you can see the number of combinations change, and we therefore need a different formula to work it out, or we need to make some modifications to our current one.  I will begin trying to find this formula, by using just one repeating letter:

AB                ABB                ABBB                ABBBB                ABBBBB        ABBBBBB

BA                BAB                BABB                BABBB                BABBBB        BABBBBB

                BBA                BBAB                BBABB                BBABBB        BBABBBB

                                BBBA                BBBAB                BBBABB        BBBABBB

                                                BBBBA                BBBBAB        BBBBABB

                                                                BBBBBA        BBBBBAB


  2                 3                    4                     5                      6                        7

= No. Of Combinations

As you can see from above, it seems that the number of combinations is just the same as the number of letters; this however is only true for one repeating letter.  When you get two repeating letters it changes quite considerably.  If you take 4 letters, as in LUCY, there are 24 combinations ordinarily, with no repeating letters.  If however there is one repeating letter as in EMMA there are only 12 combinations, if there are two letters which repeat, as in AABB, there would only be 6 combinations.  As you can see from the table below the number of combinations halve each time, but this is not the case for all different names, or combinations of letters.

Number of Letters




Number of repeated letters


1, twice

2, twice each

Number of combinations




...read more.


= (120) / (2 * 6)

= 10

(This works for this one but I will now try it for my other results to make sure that it is right.)

For XXY:

Combinations        = 3! / (2! * 1!)

                = 6 / 2

                = 3 (The correct answer)


Combinations        = 6! / (3! * 3!)

                = 720 / 36

                = 20 (Again the correct answer)

As you can see from the calculations above, the formula to work out the number of combinations for 2 repeating letters is:

Number of Combinations = Total! / (No. of the one letters! * No. of the second letters!)


C = T! / ( R! * R’!)

Where C is the total number of combinations, T is the total number of letters, R is the first repeated letters and R’ is the second repeated letters.

I have also discovered that by just adding on more R’’, and R’’’ you can work out the combinations for words or names with many repeating letters or just for all names by using the one formula.  For example I know that AABBCC has 90 combinations and I will now prove it using the formula.

Combinations         = T! / (R! * R’! * R’’!)

= 6! / (2! * 2! * 2!)

                = 720 / 8

                = 90 combinations.

I will now show how the formula can be applied to any name by incorporating all the letters, for STEVEN:

Combinations         = T! / (No. of S’s! * No. of T’s! * No. of E’s! * No. of V’s! * No. of N’s!)

                = 6! / (1! * 1! * 2! * 1! * 1!)

                = 720 / 2

                = 360 combinations of my name.

...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. Arrangements for names.

    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.

  2. Emma's Dilemma

    AACAB AACBA Total: ABAAC 20 ABACA ABCAA CAAAB ACAAB CAABA ACABA CABAA ACBAA CBAAA Results: Number of Letters: Number of different combinations: 3 1 4 4 5 20 Rule: To find the number of all the different combinations possible, from a selected number of letters ( using every letter only once, )

  1. GCSE maths coursework: Emma's dilemma

    arrangement 4 1*4 5 1*4*5 6 1*4*5*6 Can you see the pattern? so the formula for three sames numbers of a number is: a= ni/6 let's review the formula: formula for different number: a=ni formula for 2 same number: a=ni/2 formula for 3 same number: a=ni/6 Let's put them is

  2. Maths GCSE Coursework: Emma's Dilemma

    following it: 3 letters: A/A/B AAB ABA BAA 3 combinations 4 Letters: A/A/B/C AABC AACB ABCA ABAC ACBA ACAB BAAC BACA BCCA CAAB CABA CBAA 12 combinations (2 pairs of 2 different letters) I will now look at two pairs of 2 different letters with nothing following it: 4 letters:

  1. permutations & combinations

    M2 A E M1 19. M1M2 A E 20. M1M2 E A 21. M1 A M2 E 22. M1 E M2 A 23. M1 A E M2 24. M1 E A M2 Both EM1M2A and EMMA have the same spelling except for the subscripts.

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

    two letters the same quite clearly doesn't work for words with three letters the same. This seemed interesting as the formula obviously changed according to the number of repeated letters. I now wanted to find out how the formula changed according to number of the same letters.

  1. Emma’s Dilemma

    If I divide this by three because there are 3 Xs, I get the answer 5, which is how many combinations per letter. I can then add this into the formula to quickly find out haw many arrangements there are in total.

  2. Emma's Dilemma.

    of its letters the same and words with all its letters different. n n r 2x a 3=JIM NO NO 6 3=BBC YES NO 3 3=XXX NO YES 1 4=LUCY NO NO 24 4=EMMA YES NO 12 4=XXXY NO YES 4 5=RALPH NO NO 120 5=DAVID YES NO 60 5=XXXYZ

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