Arrangements for Emma
Length: 707 words (2 doublespaced pages) Rating: Red (FREE)                                  
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 4letter word I have found a pattern of 6, 12 and 24. Maybe it would be easier to see what is happening if I used larger words. What if there was a fiveletter word? How many different arrangements would there be for that? As I have found that there were 24 arrangements for a 4 letter word with all different letters and that there were 6 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 4letter 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. 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 11 22 36 424 5120 6720 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. To make it a bit easier instead of using letters as such I will use a's (any letter) and b's (any other letter). I will start with aabb: Arrangements for aabb: This is a 4letter word with 2 different. I have done this with; aabb abab baab aaba baba bbaa There are 6 arrangements. What if I had aaabb? aaabb aabab aabba ababa abaab abbaa bbaaa baaab babaa baaba There are 10 different arrangements for this sequence. What if I had an arrangement of aaaab? aaaab aaaba aabaa abaaa baaaa There are 5 different arrangements for this sequence. If I go back to aaabb; there are 3 a's and 2 b's in a total of 5 unknowns. As each letter has its own number of arrangements i.e. there were 5 beginning with a, and 5 beginning with b, I think that factorial has to be used again. Also in a 5letter word there are 120 arrangements with 24 arrangements (120 divided by 5) for each letter. As that seemed to work, I had a go at trying to work out a logical universal formula. 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 How to Cite this Page
MLA Citation:
"Arrangements for Emma." 123HelpMe.com. 17 Apr 2014 <http://www.123HelpMe.com/view.asp?id=121735>. 
