# Maths GCSE Coursework: Emma's Dilemma

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Introduction

Emma’s Dilemma

Emma and Lucy are playing with arrangements of their names. One arrangement of Lucy is:

L U C Y

A different arrangement is: Y L C U

Our task is to experiment with the different combinations of the letters in the names Lucy and Emma and find any patterns which we can use to find different combination of words and numbers. There are three parts:

Part 1:

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

Part 2:

Investigate the number of different arrangements of the letters of Emma's name.

Part 3:

Investigate the number of different arrangements of various groups of letters.

Combination of Lucy

L | U | C | Y |

LUCY | ULCY | CLUY | YLUC |

LUYC | ULYC | CLYU | YLCU |

LYCU | UCYL | CULY | YUCL |

LYUC | UCLY | CUYL | YULC |

LCUY | UYCL | CYLU | YCLU |

LCYU | UYLC | CYUL | YCUL |

Method

The method I used to find the combinations of LUCY was systematic. I knew that the name Lucy consisted of four letters, L, U C and Y, so I created a table, putting these letters in as ‘’starters’’. I then began finding the combination that go with each individual letters. I began with L. I knew that after writing the first letter (L), there were only going to be three letters left (U, C and Y), so these letters followed. After two letters were written, only two can be left. These letters came next. The next letter was the last. Now I have finished finding the combinations of L. I then used the same method to find the combination of the other ‘‘starters’’ (U, C and Y). Another way of displaying this method is shown below:

By using this method, I know that I have discovered all combinations of the word Lucy.

Middle

AEMM

EMAM

MEAM

MEAM

AEMM

EAMM

MMYE

MMEA

AMME

EAMM

MMEA

MMAE

AMEM

EMMA

MAME

MAEM

AMEM

EMAM

MAEM

MAME

AMME

There are two M’s in the name Emma, but as we are treating them as two separate letters, we can see that there are many duplicates in this table. Currently there are 24 combinations, the same as Lucy, but now we will eliminate the duplicates, leaving us with all the real combinations of Emma:

E | M | M | A |

EMMA | MEMA | MEMA | AEMM |

EMAM | MEAM | MEAM | AEMM |

EAMM | MMYE | MMEA | AMME |

EAMM | MMEA | MMAE | AMEM |

EMMA | MAME | MAEM | AMEM |

EMAM | MAEM | MAME | AMME |

Now that we have crossed out the duplicates we can create a new table:

E | M | A |

EMAM | MEMA | AEMM |

EMMA | MEAM | AMME |

EAMM | MMEA | AMEM |

MMAE | ||

MAME | ||

MAEM |

We can now see that there are actually 12 combinations in EMMA.

Analysis

- After analysing these two results, I have noticed a few things:

- The obvious thing to notice is that there are fewer combinations in the word Emma because of the fact there are two M’s in it. This means that many combinations are cut out of the total count, as shown above.

- Taking this idea further, we can also see that the amounts of combinations in Emma are exactly half the amount of Lucy. However, this analysis could be mathematical or incidental, we do not know for sure, but it something to watch out for.

Using this methodical method is ok for words such as Lucy and Emma, but in another case such as a six-letter word, it would take very long and be extremely inconvenient. Therefore, a method must be found to make this process quicker and easier.

I am going to do this by experimenting with combinations of different amounts of letters, and see if any pattern emerges. I will use 1, 2, 3 and 4 letters, and put the results in a table.

1 letter: A

A: 1 combination (1x1)

2 letters: A/B

AB,BA: 2 combinations (2x1)

3 letters: A/B/C

A | (BC) |

B | (AC) |

C | (AB) |

Here we have three starting letters (A,B and C). We can do this systematically. We start with A. Now, because we have written A, we know that there can only be two more letters which go next, B and C. We also know from our previous case that with 2 letters, there are 2 combinations. We can also do this with the other two starting letters (B and C). So now we have three starting letters with two other letters following them (which can be rearranged). So in conclusion, we have 3 starting letters and 2 combinations for each one. So it’s 3x2x1=6 combinations

All letters in brackets can be rearranged

4 letters: A/B/C/D

A | (BCD) |

B | (ACD) |

C | (ABD) |

D | (ABC) |

The same method is required her, except now there is another letter (D). But we still follow the same procedures. There are four starting letters (A,B,C and D), and three letters which go with them. And from our last case we already know that with three letters, there are 6 combinations (3x2x1). So now, we just multiply this by 4

4x3x2x1 (or 4 multiplied by 6) =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 |

Conclusion

1 letter repeated 3 times:

A/A/A

AAA 1 combinations

Now we will put something in front of this, and see if anything changes:

1 letter repeated 3 times + 1 different letter:

A/A/A/B

AAAB

AABA

ABAA

BAAA 4 combinations

I used the same systematic method as I did for all the other cases as I did for this, but when I had finished it, I saw something quite interesting. Because there is only one different letter in this (B), all that is really happening is the B is moving along the letters (highlighted in BLUE).

I will now look at a case where two letters are after of the repeated letters:

1 letter repeated 3 times + 2 different letters:

A/A/A/B/C

AAABC

AAACB

AACBA

AACAB

AABCA

AABAC

ABAAC

ABACA

ABCAA

ACAAB

ACABA

ACBAA

Next, we will be looking at one letter being repeated more than three times. We will look at one letter being repeated four times, with nothing after it, and then with something after.

1 letter repeated 4 times:

A/A/A/A

AAAA 1 combination

1 letter repeated 4 times + 1 different letter:

A/A/A/A/B

AAAB

AABA

ABAA

BAAA 4 combinations

We will now move on to looking at one letter bring repeated five times:

1 letter repeated 5 times:

A/A/A/A/A

AAAAA 1combination

1 letter repeated 5 times + 1 different letter:

A/A/A/A/A/B

AAAAB

AAABA

AABAA

ABAAA

BAAAA 5 combinations

My results are shown in the table below:

Number of letters | All different (ABCD) | 1 letter repeated twice | 1 letter repeated 3 times | 1 letter repeated 4 times | 1 letter repeated 5 times |

1 | 1 | ||||

2 | 2 | 1 | |||

3 | 6 | 3 | 1 | ||

4 | 24 | 12 | 4 | 1 | |

5 | 120 | 60 | 20 | 4 | 1 |

6 | 720 | 360 | 5 | ||

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

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