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Emma’s Dilemma

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Introduction

Catherine Brown Emma's Dilemma During the course of this investigation I hope to find a formula that gives Emma the solution to her problem: how many permutations there are of arranging a given combination of letters. A permutation is different ways of arranging a set combination. A combination is a given set of numbers/ letters/ objects were order doesn't matter. I have decided to start with a 3-lettered word, all different. Where the results I gain from this investigation will help me to predict the next set of results accurately with evidence to back up my claims. My 3-lettered word will be AMY. This is 3 letters, none of them the same. 1. AMY 2. MAY 3. YAM 4. MYA 5. AYM 6. YMA Arranging the three letters has a possible 6 permutations. This could also be written as 3*2*1. I have chosen to write the formula in this format, instead of 3*2 (if it is multiplied by 1 the answer is till the same) as there are 3 letters in the word This time the 3-lettered word shall have 2 letters the same so I can find out if this will affect the amount of permutations of the given letters. In this case the Y has been substituted by a second M. 1. AMM 2. MAM 3. MMA Arranging 3 letters with 2 the same gives you 3 permutations. This is half the original value as some of the permutations repeat themselves. ...read more.

Middle

60480 9*8....*1/(3*2*1) 10 1814400 10*9....*1/(2*1) 604800 10*9....*1/(3*2*1) n n*(n-1)*(n-2)...*1/(2*1) n*(n-1)*(n-2)...*1/(3*2*1) The table shows that if there are 2 letters the same then the formula ends in 2*1. If there are 3 letters the same then the formula ends in 3*2*1. So the formula for a letter being used more than once is n*(n-1)*(n-2)*(n-3)....*1/m*(m-1)*(m-2)*(m-3)...*1. In this formula m is how many times a letter is used. This is how the formula works: if you have a 5 lettered word with 3 letters the same: 5*(5-1)*(5-2)*(5-3)*(5-4)/3*(3-1)*(3-2) which is 20. To prove this here are all of the possible permutations for a 5-lettered word, 2 the same. 1. GAGOG 2. GAGGO 3. GAOGG 4. GOAGG 5. GOGAG 6. GOGGA 7. GGOGA 8. GGOAG 9. GGAOG 10. GGAGO 11. GGGAO 12. GGGOA 13. AGOGG 14. AGGOG 15. AGGGO 16. AOGGG 17. OGAGG 18. OGGAG 19. OGGGA 20. OAGGG This proves that the formula is correct for words with a letter repeated. This formula will only work if there is one set of letters the same; if there are more then a different formula is needed. This time I will look at a 4-lettered word consisting of 2 letters the same and 2 letters the same. The 4-lettered word that will be used is AABB. 1. AABB 2. ABBA 3. ABAB 4. BBAA 5. BABA 6. BAAB There are a total of 6 permutations for this word, which could be written as (4*3*2*1) /((2*1)*(2*1)) as there are 4 letters in the word and there are 2 letters repeated twice. ...read more.

Conclusion

If there are 2 sets of repeated letters then there are 2 sets of brackets. So using this pattern if there were 3 sets of repeated letters there would be 3 sets of brackets, and if there were 7 sets of repeated letters then there would be 7 brackets. So this gives the formula n*(n-1)*(n-2...*1/((m*(m-1)*(m-2)...*1)*(p*(p-1)*(p-2))...*1 where p is the second set of repeated letters. For example if there was a 7-lettered word with 2 letters the same and 3 letters the same then the formula would be 7*6*5...*1/((2*1)*(3*2*1)) which is 420 permutations. To show this I am going to use the word MATHEMATICS. In this word there are 3 sets of repeated letters so there will be 3 divisions. This means that the formula will be 11*10*9...*1/((2*1)*(2*1)*(2*1)). This means that there are a possible 4989600 permutations for an 11-lettered word with 3 sets of repeated letters. The conclusion to this investigation is that I have found that however many sets of repeated letters there are in a word gives the amount of brackets in the formula. For example if there are 3 sets of repeated letters then there are 3 brackets. If there are 10 sets of repeated letters there will be 10 sets of brackets etc. if the word has 15 letters i.e. 5 the same and 10 the same then the formula still has the 15*14*13...*1, which is 1307674368000. This is then divided by 5*4...*1, then by 10*9...*1 which gives a total of 3003 permutations. Page 1 ...read more.

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