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Emma's Dillemma - Rearranging Emma's Name in different permutations

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Introduction

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 that are all different. That's twice as many as EMMA, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For instance 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 anna: anna anan aann nana naan nnaa There are 6 arrangements for aabb. From 2 different letters to all different letters in a 4-letter word I have found a pattern of 6, 12 and 24. ...read more.

Middle

So as a general formula for names with x number of letters all different I have come up with a formula. With Lucy's name; 1x2x3x4 = 24. With qlucy; 1x2x3x4x5 = 120. This is called a factorial. There is a button on most scientific calculators which basically does this formula for you according to the number of letters in the word. If I key in (assuming all the letters are different) factorial 6, I get it gives me 720, which makes sense because 720 divided by 6 equals 120 which was the number of arrangements for a 5 letter word, and it continues to follow 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 any letters are repeated? Like in Emma; it has 4 letters but 2 of which are the same. 4 factorial equals 24, but I could only find 12, which means that there must be more to it that just factorial in that way. ...read more.

Conclusion

I came up with; The total number of letters factorial, divided by the number of a's, b's ect factorised and multiplied Formula for emma= 4!/1!x1!x2!=48 /4=12 For example: A five letter word like aaabb; this has 3 a's and 2 b's. So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10 A four letter word like aabb; this has 2 a's and 2 b's So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6 A five letter word like aaaab; this has 4 a's and 1 b 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 (all different!) Number of A's Number of B's Number of arrangements 1 1 0 1 2 - - 2 3 - - 6 4 - - 24 5 - - 120 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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