French Holidays
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Emma's Dilemma
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Introduction
Emma’s Dilemma
Emma and Lucy are playing with arrangements of the letters of their names.
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.
Emma’s Dilemma
Emma and Lucy were playing with arrangements of the letters of their names.
Part 1
My first task was to investigate the number of different arrangements for the letters in Lucy’s name. I began by listing the arrangements, one by one, following a certain structure (a structure which will hereafter be used in similar situations throughout the rest of this investigation). This started with me dividing the name “Lucy” up into its individual parts i.e. the letters L, U, C and Y. By re-arranging these, I was then able to find new arrangements. Starting with LUCY (obviously), I moved on to find LUYC (by exchanging the two letters at the end). When I found that there were no longer any arrangements starting with LU, I simply replaced the second letter with the subsequent letter and started to look for arrangements starting with LC. When the arrangements starting with L had run out altogether, I took the next letter i.e. U and started to look for arrangements starting with that. In the end, I found that there were 24 different arrangements for the letters in Lucy’s name:
LUCY | ULCY | CLUY | YLUC |
LUYC | ULYC | CLYU | YLCU |
LCUY | UCLY | CULY | YULC |
LCYU | UCYL | CUYL | YUCL |
LYUC | UYLC | CYLU | YCLU |
LYCU | UYCL | CYUL | YCUL |
Middle
EAMM
MMEA
AMME
MMAE
MAEM
MAME
I discovered 12 different arrangements in which the two M’s in Emma’s name could be exchanged and 24 in which they were individual. This tells me that the prior was half of the latter. Looking at my results now, this is quite understandable. I can see that for every two arrangements in which the M’s were separate, there was one of the opposite. For example, EM1M2Aand EM2M1A could just as easily have been EMMA.
However, to verify the results, I decided to have a look at a similar name “Zeek”. Again, I found 24 different combinations where the M’s were identifiable and 12 where they weren’t so there shouldn’t be a problem:
ZE1E2K | E1ZE2K | E2ZE1K | KZE1E2 |
ZE1KE2 | E1ZKE2 | E2ZKE1 | KZE2E1 |
ZE2E1K | E1E2ZK | E2E1ZK | KE1ZE2 |
ZE2KE1 | E1E2KZ | E2E1KZ | KE1E2Z |
ZKE1E2 | E1KZE2 | E2KZE1 | KE2ZE1 |
ZKE2E1 | E1KE2Z | E2KE1Z | KE2E1Z |
ZEEK | EZEK | KZEE |
ZEKE | EZKE | KEZE |
ZKEE | EEZK | KEEZ |
EEKZ | ||
EKZE | ||
EKEZ |
I then decided to see if there was a trend between the number of repetitions there were in a word, and the number of different arrangements, in which those repetitions were interchangeable, that the letters in that word could form. In order to do this, I kept the length of the words constant (i.e. at four characters per word) so as it would not interfere with the results I got.
I decided to look at the word “baaa”. I found that, as with the names “Emma” and “Zeek” before it, the letters in it could appear in 24 different arrangements where the repeated letters were separable. However, there were only 4 different arrangements in which they were not:
BA1A2A3 | A1BA2A3 | A2BA1A3 | A3BA1A2 |
BA1A3A2 | A1BA3A2 | A2BA3A1 | A3BA2A1 |
BA2A1A3 | A1A2BA3 | A2A1BA3 | A3A1BA2 |
BA2A3A1 | A1A2A3B | A2A1A3B | A3A1A2B |
BA3A1A2 | A1A3BA2 | A2A3BA1 | A3A2BA1 |
BA3A2A1 | A1A3A2B | A2A3A1B | A3A2A1B |
BAAA | ABAA | |
AABA | ||
AAAB |
Again, to credit my findings, I investigated a similar word “mooo”. And, again, I found that the letters it is spelt of could fit in 24 different arrangements where the repetitions of the letter O could be identified and 4 where they couldn’t:
MO1O2O3 | O1MO2O3 | O2MO1O3 | O3MO1O2 |
MO1O3O2 | O1MO3O2 | O2MO3O1 | O3MO2O1 |
MO2O1O3 | O1O2MO3 | O2O1MO3 | O3O1MO2 |
MO2O3O1 | O1O2O3M | O2O1O3M | O3O1O2M |
MO3O1O2 | O1O3MO2 | O2O3MO1 | O3O2MO1 |
MO3O2O1 | O1O3O2M | O2O3O1M | O3O2O1M |
MOOO | OMOO | |
OOMO | ||
OOOM |
Conclusion
D1E1E2D2 | E1D1E2D2 | E2D1E1D2 | D2D1E1E2 |
D1E1D2E2 | E1D1D2E2 | E2D1D2E1 | D2D1E2E1 |
D1E2E1D2 | E1E2D1D2 | E2E1D1D2 | D2E1D1E2 |
D1E2D2E1 | E1E2D2D1 | E2E1D2D1 | D2E1E2D1 |
D1D2E1E2 | E1D2D1E2 | E2D2D1E1 | D2E2D1E1 |
D1D2E2E1 | E1D2E2D1 | E2D2E1D1 | D2E2E1D1 |
DEED | EDED | ||
DEDE | EDDE | ||
DEED | EEDD |
The formula “the number of different arrangements for the letters in a name = n!/a!” obviously didn’t apply in this case as there were two completely unrelated repetitions. Therefore:
The number of different arrangements for the letters in a name = n!/a!×b!
N.B. b referring to the number of repetitions of the second variety.
The b! works on the same principle as the a! before it; that, just as if they were words themselves, the duplicate letters can be arranged in a number of different arrangements, taking into account that they are distinguishable, according to the number of them there were. Hence, the number of different arrangements for the letters in a name is equal to n! divided by a! b! or a! multiplied by b!. If there were more repeated letters, this could be expanded. c would be used to represent the number of times the third letter is repeated and d for the next and so on.
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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