# emmas dilema

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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

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.

Part 1

Set 1 | Set 2 | Set 3 | Set 4 |

LUCY | UCYL | CUYL | YCUL |

LUYC | UCLY | CULY | YCLU |

LCUY | ULCY | CLUY | YLUC |

LCYU | ULYC | CLYU | YLCU |

LYCU | UYCL | CYUL | YULC |

LYUC | UYLC | CYLU | YUCL |

I have investigated the number of arrangements of the letters of Lucy’s name in an organized way. I have written them out in sets with Set 1 beginning with L. Set 2 with U, Set 3 with C and Set 4 with Y. in each set the arrangements are written out in an order i.e. in set 1 the first 2 arrangements begin with LU, the 2 after that begin with LC and the 2 after that begin with LY. There are 24 different arrangements.

Examples

If it is just one letter then you cannot change it. If it is two letters like AB then it can be changed into AB or BA and there are 2 arrangements.

If there are three letters like ABC then there are 6 arrangements. I organized this so that set 1 begins with A, set 2 with B and set 3 with C.

Set 1 | Set 2 | Set 3 |

ABC | BAC | CAB |

ACB | BCA | CBA |

If there are four letters like LUCY then there are 24 arrangements as shown at the beginning of part 1.

Results Table

Number of letters | Arrangements |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

I constructed a table to see if there were any connections between the number of letters and the different arrangements. Looking at the table of results I can see that there is a connection between the two columns.

I have found that 4!, works in finding the number of arrangements for Lucy’s name.

4! = (4x3x2x1) = 24.

Using 4!, works because the word Lucy has 4 letters and this means there are 4 letters to start an arrangement with L, U, C or Y. After the first letter is picked there are 3 other letters to choose from e.g. if L was picked then only U, C and Y are left. 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. LUC is written and Y is left.

Remembering that 4! means (4x3x2x1) the 4 stands for the number of letters you can begin with, the 3 is the number of letters available after the first letter is written. The 2 is the number of letters available after the second letter is written. The 1 is the last letter that can be picked after three letters have been written.

The formula for finding the number of arrangements for a word with any number of letters but without any repeated letters is n! and the n is the number of letters in a word. The n! means that there are n letters to begin with. For the second letter there are n-1 letters to choose. For the third letter there will be n-2 letters to choose. For the forth letter there will be n-3 letters to choose and so on.

So the equation would be n x(n-1) x(n-2) x(n-3) and so on. This is how n! works.

Part 2

These are all the different arrangements of letters of Emma’s name and it has 4 letters with 2 repeated letters.. I have written them out in an order so it is clear. Set1 begins with E, set2 begins with M and set3 begins with A. There are 12 arrangements of her name.

Table 1

Set 1 | Set 2 | Set 3 | |

EMMA | MMAE | AMME | |

EMAM | MMEA | AMEM | |

EAMM | MAEM | AEMM | |

MAME | |||

MEMA | |||

MEAM |

Middle

M2M1EA

AM2EM1

EAM1M2

M1AEM2

AEM1M2

EAM2M1

M2AEM1

AEM2M1

M1AM2E

M2AM1E

M1EM2A

M2EM1A

M1EAM2

M2EAM1

EMMA has 4 letters in her name so it would be expected that she has 24 arrangements because 4! is 24 (the 4! as explained in Lucy) but this is not the case because she has two Ms in her name. If the two Ms in Emma’s name were switched then this would make no difference to the number of arrangements as the name would be the same. So there are only 12 arrangements due to the fact that there are two Ms in the name. The 2 Ms have 2! arrangements. For each arrangement there is another the same with the Ms reversed.

In table 2 I have arranged them so that the corresponding arrangement can be clearly seen on the other side. Sets 1-3 have the M1 in front of the M2 and sets 4-6 have the M2 in front of the M1. There are 24 arrangements and 12 identical pairs. This means that there is half the number of arrangements for a word with 2 repeated letters.

Using the results I can say that the formula for finding out the number of arrangements of Emma is 4!/2 This means (4x3x2x1)/(2x1) = 24/2 = 12

Using the evidence gathered so far the formula might be n! divided by 2 if there are 2 repeated letters. The unique arrangements will only be half the total number of arrangements i.e. 24/2!

The problem with this is that the formula would not work for a 5 letter word with 5 repeated letters because 5!/5 = 24 as there is only one arrangement. This means that this formula is not the complete thing yet.

Examples

I am going to see if I can find the formula which lets me know the arrangements of a word with a letter repeated more than once based upon my results for Emma’s arrangements. I will use further examples carry this out.

AAAAB is a 5 letter word with 4 repeated As and this will be used as an example.

Set 1 | Set 2 |

AAAAB | BAAAA |

AAABA | |

AABAA | |

ABAAA |

Again I have put the results in a table but it is impractical to make a table like the second one for Emma because there are will be 120 arrangements. Set 1 begins with A and set 2 begins with B. I have shown the arrangements of AAAAB when the As are all treated the same . With Emma’s name there are 2 Ms which meant that the number of arrangements was divided by 2. I have found that there are 5 different arrangements for AAAAB. This means that the formula for AAAAB is n! divided by 24. This can be put simpler by saying 5!/4! = 5.

It can be seen how 120 is divided by 24 because the As when written fully to give 120 arrangements would be sorted out with 4 identical arrangements.

If the As were taken out and labelled A1A2A3A4 and treated as a separate word it would have 24 different combinations because of 4! This is why 120 is divided by 24 in 5!/4!

If all letters are the same in a 5 letter word then obviously you cannot change it. If there are 4 letters repeated such as AAAAB as shown in the previous example then there are 5 arrangments. If there are 3 repeated letters in a 5 letter word then there are 20 examples. For example AAABC has 3 repeated letters and has 5 letters. There are 20 different arrangements.

Set 1 | Set 2 | Set 3 |

AAABC | BAAAC | CAAAB |

AAACB | BACAA | CAABA |

AABCA | BAACA | CABAA |

AABAC | BCAAA | CBAAA |

AACAB | ||

AACBA | ||

ABAAC | ||

ABACA | ||

ABCAA | ||

ACAAB | ||

ACABA | ||

ACBAA |

Set 1 | Set 2 |

ABBB | BBBA |

BBAB | |

BABB |

Conclusion

Examples

AABB has 4 letters with 2 repeated As and 2 repeated Bs.

Set 1 | Set 2 |

AABB | BBAA |

ABAB | BABA |

ABBA | BAAB |

This is the table showing me that there are 6 different arrangements of AABB.

I can prove this by doing 4!/ (2! x 2!)

AABBCC has 6 letters with 2 repeated As, 2 repeated Bs and 2 repeated Cs.

I will not do a results table for AABBCC because it would be impractical as I have found out that doing 6!/ (2! x 2! x 2!) = 90 arrangements

The reason for this is because the 2 As have 2! Arrangements, the 2 Bs have 2! arrangement and the 2 Cs have 2! arrangements.

I will now do this with other words and arrangements to prove that the formula works.

Word | Formula | Number of arrangements |

BANANA | 6!/ (2! x 3!) | 60 |

AABBCCDD | 8!/ (2! X 2! X 2! X 2!) | 2520 |

OSMOSIS | 7!/ (2! X 3!) | 420 |

ONOMATOPOEIA | 12!/ (4! X 2!) | 9979200 |

REMEMBER | 8!/ (2! X 3! X 2!) | 1689 |

After doing this table the formula for finding the arrangements of a word with any length, any number of repetitions and any number of repeated letters is n!/ (a! x b! x c! and so on.)

A, B, and C represent each repeated the number of times a single letter is repeated.

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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