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Emma’s Dilemma

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Introduction

Mathan Navaratnam 11F 24.10.2001 Mathematics Coursework Emma's Dilemma The different arrangements for the name Emma are as follows: Emma emam eamm mmae Mmea meam mema mame Maem amme amem aemm There were a total of twelve combinations of the rearrangement of the name Emma. The name Emma consists of four letters where three of them being different, the two like letters being m. Emma's friend however, Lucy, name consists again of four letters but this time all the letters are different. The number of arrangements for the name Lucy is as follows: Lucy ucyl cylu ycul Luyc ucly cyul yclu Lcuy ulcy culy yulc Lcyu ulyc cuyl yucl Lyuc uycl clyu ylcu Lycu uylc cluy yluc This time there were twenty-four different possibilities in rearranging the name Lucy. This is significant, as it is exactly double the number of arrangements of the name Emma where there were 4 letters but where only 3 were different. Looking back at the table above I think it would be useful to note how to predict the number of arrangements. So far I have worked out for names with four letters. ...read more.

Middle

This is interesting as for one letter there is one combination (1), for two letters there is two combinations (1 X 2), for three combinations there were six combinations, (1 X 2 X 3), for four letters there were 24 combinations (1 X 3 X 4) and for five letters there were 120 combinations (1 X 2 X 3 X 4 X 5). That means by following this pattern, for a six letter word where all the letters are different there should be 720 combinations, (1 X 2 X 3 X 4 X 5 X 6). This would make sense as there would be six starting letters and the remaining five different letters can be rearranged 120 times, as we discovered earlier. So for each starting letter there will be 120 combinations and there are 6 different starting letters so there will be 120 X 6 = 720 combinations. 720 = (1 X 2 X 3 X 4 X 5 X 6) which follows the formula which I found earlier. (1 X 2 X 3 X 4 X 5 (the formula for five different letters) X 6 (number of different stating letters). This is expressed as factorial or !. Factorial represents the calculation of 1x2x3x4x5x6x7... etc. ...read more.

Conclusion

There were 3 x's and 3! = 6. There were 2 y's and 2! = 2. There were a total of five letters in the word a 5! = 120. From this we can deduce that 3!x2! = 12. This is significant as 120 (which is 5!) divided by 12 gives us 10 which is exactly the number of combinations there was for xxxyy. Therefore 5! / 3!2! = 12. This clearly works in the case of xxxyy. We can replace numbers in this formula to letters. Therefore the formula would be n!/x!y! To test whether this is the correct formula I will apply it to the above examples. So for xxxyy; there are 3 x's present and 2 y's present. Therefore 1x2x3x4x5 / (1x2x3) x (1x2) = 120/12 = 10. For xxxxy, there are 4 x's present and 1 y present. Therefore 1x2x3x4x5 / (1x2x3x4) x (1) = 120 / 24 = 5. For xxyy there are 2 x's present and 2 y's present. Therefore 1x2x3x4 / (1x2) x (1x2) = 24 / 4 = 6 This formula should also work where every letter of the word is different. So for a word such as vwxyz there is only one of each letter present. Therefore 1x2x3x4x5 / (1)x(1)x(1)x(1)x(1) = 120 / 1 = 120. The above calculations prove that my formula is correct. ...read more.

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