# Emma's Dilemma - Rearranging Emma's Name in different permutations

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Introduction

Emma's Dilemma - Rearranging Emma's Name in different permutations I am going to investigate the number of different arrangements for the letters in the name EMMA. The name EMMA consists of 2 different letters and 2 letters the same. I am going to investigate other names with the same combination of letters; I will try and find a formula for these combinations. 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. I will then look at the name LUCY. This combination consists of all the letters being different. I will also try and find a formula for this arrangement, and number of letters. 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. ...read more.

Middle

combinations beginning with each of the letters in the word I predict that there will be 120 arrangements for qlucy, 24 for q, 24 for l, 24 for u; and so on. 120 divided by 6 (number of letters) equals 24. Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter. For the sake of convenience I have used lucy and put a q in front of it to show that there are 24 different possibilities with each letter of a 5 lettered name being all different. qlucy qucyl qcylu qycul qluyc qucly qcyul qyclu qlcuy qulcy qculy qyulc qlcyu qulyc qcuyl qyucl qlyuc quycl qclyu qylcu qlycu quylc qcluy qyluc I can see that the numbers of different arrangements are going to dramatically increase as more different letters are used. 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. ...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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