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

different letters to all different letters in a 4-letter 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 five-letter 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 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 qyculqluyc qucly qcyul qycluqlcuy qulcy qculy qyulcqlcyu qulyc qcuyl qyuclqlyuc quycl qclyu qylcuqlycu quylc qcluy qylucI 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 Arrangements1-12-23-64-245-1206-720So 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 4-letter word with 2 different. I have done this with;aabb abab baabaaba baba bbaaThere are 6 arrangements. What if I had aaabb? aaabb aabab aabba ababa abaababbaa bbaaa baaab babaa baabaThere are 10 different arrangements for this sequence.What if I had an arrangement of aaaab? aaaab aaaba aabaaabaaa baaaaThere 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 5-letter 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 multipliedFormula for emma=4!/1!x1!x2!=48 /4=12For example:A five letter word like aaabb; this has 3 a's and 2 b's. So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10A four letter word like aabb; this has 2 a's and 2 b's So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6A five letter word like aaaab; this has 4 a's and 1 b So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5Five letter words like abcde; this has 1 of each letter (no letters the same)So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24All these have been proved in previous arrangements. This shows that my formula works.  Total letters(all different!)Number of A’sNumber of B’sNumber of arrangements11012--23--64--245--120321342265231062415330143145321063320