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Emma's Dilemma

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Introduction

Emma's Dilemma Arrangements for Emma: emma emam eamm mmae mmea meam mema mame maem amme amem aemm There are 12 possibilities; note that there are 4 total letters and 3 different. What if they were all different like Lucy? Arrangements for Lucy: lucy ucyl cylu ycul luyc ucly cyul yclu lcuy ulcy culy yulc lcyu ulyc cuyl yucl lyuc uycl clyu ylcu lycu uylc cluy yluc There are 24 different possibilities in this arrangement of 4 letters all different. Double the amount as before with Emma's name, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For example there are 6 arrangements with Lucy beginning with L, and 6 beginning with u and so on. 6 X 4 (the amount of letters) gives 24. What if there were 4 letters with 2 different? Arrangements for aabb: aabb abab baab abba baba bbaa There are 6 arrangements for aabb. ...read more.

Middle

With Lucy's name; 1x2x3x4 = 24. With qlucy; 1x2x3x4x5 = 120. However this is expressed as factorial. There is a button on most scientific calculators with have embedded this factorial button feature generally sowing as an exclamation mark. All it does is save the time of having to put in to the calculator 1x2x3x4x5x6x7..... Ect. You just put in the number and press factorial and it will do 1x2x3... until it get to the number you put in. If I key in (lets say the number of letters all different) factorial 6 I get it gives me 720, with makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word and it continues to fall in that pattern. Total Letters (all different) Number of Arrangements 1 1 2 2 3 6 4 24 5 120 6 720 So now that I've explained the pattern of general x lettered words, what do I do if there are repeat letters? ...read more.

Conclusion

Formula = For example: A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's). So : 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10 !!! A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's) So : 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6 !!! A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y) So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5 !!! Five letter words like abcde; this has 1 of each letter (no letters the same) So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24 All these have been proved in previous arrangements. This shows that my formula works !!! Total Letters Number Of X's Number Of Y's Number of Arrangements 3 2 1 3 4 2 2 6 5 2 3 10 6 2 4 15 3 3 0 1 4 3 1 4 5 3 2 10 6 3 3 20 ...read more.

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