Emma’s Dilemma maths coursework - page 1
Keywords: Emma’s Dilemma maths coursework GCSE
By slashwk on 23/11/2006 17:25:42
Level: GCSE Key Stage 4 (Years 10-11)
Page Number: 1 of 6 pages: 1 2 3 4 5 6
Emma is experimenting with the arrangement of letters in her name:
One arrangement is: E M M A
A different arrangement is: M E A M
Another arrangement is: A E M M
To help me get an understanding of how the arrangement permutations work, I am firstly going to investigate the t types of arrangements with simpler words in order to find any correlations, and possibly find a rule which I could apply to any circumstance and achieve a true consequence every time. I will use simple letters (A,B,C) to keep it easy to understand.
Here I am experimenting with one letter:
1. A
I can see that there is obviously only one combination with just one letter.
Here I am experimenting with two letters:
1. AB
2. BA
There are only two combinations with two letters (could there possibly be the obvious pattern of: n letters = n combinations?)
Here I am experimenting with three letters:
1. ABC
2. ACB
3. BAC
4. BCA
5. CAB
6. CBA
This is interesting as there are six combinations with three letters (so there isn’t an n letters = n combinations relationship, better keep investigating!)
Here I am experimenting with four letters:
1. ABCD 7. BACD 13. CABD 19. DABC
2. ABDC 8. BADC 14. CADB 20. DACB
3. ACDB 9. BCAD 15. CBAD 21. DBAC
4. ACBD 10. BCDA 16. CBDA 22. DBCA
5. ADCB 11. BDAC 17. CDBA 23. DCAB
6. ADBC 12. BDCA 18. CDAB 24. DCBA
This is very fascinating as there are now twenty-four different combinations when there are four different letters. The combination numbers are getting quite large now, so I’m going to create a table from my acquired results to see whether I can draw any conclusions and possibly figure out a formula which satisfies any number of letters.
Number of letters (n) 1 2 3 4
Number of combinations (c) 1 2 6 24
Maybe I’ll be able to create a formula with the nth term. In order to do this I have to figure out the differences between the combination numbers:
Number of letters (n) 1 2 3 4
Number of combinations (c) 1 2 6 24
1st difference 1 4 18
2nd difference 3 14
3rd difference 11
Obviously there isn’t any correlation in regards to the nth term.
I have, however, noticed a different noticeable pattern. The number of combinations that a number of letters will create is the number of combinations which were possible for the last term, multiplied by the number of letters in the particular term you are finding the number of combinations for. For example:
Number of letters (n) 1
2
3
4
Number of combinations (c) 1 2 6 24
(The previous number of combinations multiplied by n gives the number of combinations for n)
This correlation shows us that in order to find out how many combinations you could possibly have with four different letters, you would have to times 1 x 2 x 3 x 4. When you times numbers like
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