permutation alogrithm?
478539
Hmmmmm. \nPermutation algorithm is the process of choosing r things out of n possible things, where r <= n.\nMathematical formula is :\nn! / (n -r)! \n\nNote that each of the collection of r things are considered different EVEN in order in which they are chosen.\n\nFor example, for objects ,  a,a,b,b,c,c,d,d. We have n =8.\nSuppose r = 3. Then, these collections ( a,b,c ) and \n( a,c,b ) are considered different because order is important here. Note that ( a,b,c ) here denotes that the case where a is chosen first, b second, and c third.\n\nIts counterpart is the concept of combination of things as opposed to pemutation of things mentioned above. In the combination concept, using the example above, the collections ( a,b,c ) and ( a,c,b ) are considered same because order is NOT important here. \n\nThus permutation counts the number of ways r things can be chosen from a total of n things with regard to order in which the objects are selected. \n\nOn the other hand, in like manner one can define combination as one counts the number of ways r things can be chosen from a total of n things without regard for the order in which the objects are selected. The mathematical formula here is \nn!/ [ (n -r)! r! ]
706198
5*4*3 = 60\n\nExample:\nNow let's suppose we have 10 letters and want to make groupings of 4 letters. It's harder to list all those permutations. To find the number of four-letter permutations that we can make from 10 letters without repeated letters (10_P_4), we'd like to have a formula because there are 5040 such permutations and we don't want to write them all out! \n\nFor four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040. This is part of a factorial \nhttp://mathforum.org/dr.math/faq/faq.comb.perm.html
20923
the formula of permutation of 2 items is n(n-1)\nthe formula of combinations of 2 items is 0.5n(n-1)\nso u see easily that the first is the double of the second
868097
-\naside from 1,\n2,3,4,6,12
943885
if you take the numbers one-by-one: 25! / ( (25-1)!*1! )\n\ncombination from n numbers taken by k:  n! / ( (n-k)!*k! )\n\nfor example:\n    -25 numbers, taken by 7: 25! / (18!*7!)\n    -25 numbers, taken by 4: 25!/ (21!*4!)
57033
the answer is 3 because a-b|a-b & a-b|a so a-b|b
522619
1 1 1 2\n1 1 2 1\n1 2 1 1\n2 1 1 1\n\nNo other case is possible, So n(A)=4\n\n\n1 1 1 3\n1 1 3 1\n1 3 1 1\n3 1 1 1\n\n1 1 2 2\n1 2 1 2\n1 2 2 1\n\n2 1 1 2\n2 1 2 1\n2 2 1 1\n\nNo other possibility\nn(B)=10
695512
Answers choices are:\n     1, 4, 3, 2, 2,\n      3, 1, 4, 2, 1,\n      1, 1, 4, 2, 1\n      3, 2, 3, 4, 1\n      3, 4, 1, 2, 4
665944
I am giving you the pseudocode. You need to implement it yourself.\nAssume the notation [abc] means allowed permutations of string 'abc'\nHence  [abc] = a[bc] + [bc]a\nThe most trivial case is [a] = a\nThis logic lends itself to a recursive routine that should be implementable in each of the languages you mentioned.\nIncase you cant, tell me, I might look into it! :-)
936395
Its called Counting Principal\n\nFactorial- the product of a number and all of the numbers less than the number all the way to 1. \n\n0!= 1\n\nAn arrangement in which order is important is called a permutation.\n\nIf you have n objects taken r at a time, the formula for a permutation is: nPr= n!/(n-r)!\n\nA word processing program requires a user to enter a 7-digit registration code made up of the digits 1,2,4,6,15,67, and 9. Each number has to be used more than once. How many differnt registration codes are possible?\n\nn=7\nr=7\n         7P7= 7!/(7-7)!= 7!/7= 5040\n\nSo basically 7*6*5*4*3*2*1= 5040\n\nHope I could help!\n\nP.S You need a scientific calculator to do this!
270029
There are of course good ways and bad ways of doing this.  The best way involves using a search tree like DFS and then pruning.  Instead of showing you something this complex (since I am unsure of your level of understanding) I will show you a common simple algorithm.  I will leave the code up to you, here is the psuedo-code:\n\n----------------------------------------------------\nALGORITHM JohnsonTrotter(n)\n//input: positive integer n\n//output: a list of all permutations of (1...n)\n\ninitialize the first permutation with (1,2...n) and with a second correspoing list ('L','L',...,'L').\n\n//Definition: mobile integer = if the corresponding letter is L \n//and the left adjacent number is smaller or vise versa.\n \nwhile there exists a mobile integer k do\n     find the largest integer K\n     swap K and the adjacent integer (left one if L) right otherwise\n     reverse the direction of all integers that are larger than k\nend do while\n----------------------------------------------------\nthis is really a simple algorithm to follow and to code, I am working off memmory here so it may not be the exact terms used but the algorithm is 100% I assure you.  Hope this helps and have fun.
860428
u gotta use somesort of code
438115
Okay, first you have to calculate how many the combinations of the first three teams are. At this point, arrangement is irrelevant. So, it'll be 15C3. Sorry, I don't exactly remember the full formula. Then, you'll get 455. Now, the arrangement is becoming relevant, since, the three teams could be number one, two, or three. So, at this point, permutation takes place. You are to multiply the previous answer with 3P3(or 3!, same things, aren't they?!). So,the answer will be 2,730! I know, it could be similar to another or other answer(s), but in math, the way of solving a question is more important than the answer itself! (That's what i think!)
1031756
It would help if you used some parentheses to indicate which terms are being divided into which terms!  I suspect that just using the order of precedence for operations will not result in the problem you are actually trying to solve.
939247
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
588383
Since to=o^5t, every element of <t,o> can be written in the form o^nt^m. Since t has order 2 and o has order 12, this gives 24 elements possible. Since H has elements that are odd permutations, H intersect A7 will have index 2 in H, so the collection of even permutations in H will have 12 elements as will the collection of odd permutations.
993077
not that I know...you might have to wright your own algorithm!\n\nouch...have fun!\n\nbut check links!!