mathematics permutation?
360115
2-M 2-A 2-T Therefore the term in the denominator. Wish you could've posted the exact question. I think -1 has been used to mean rearrangement ie. In what other ways in which MATHEMATICS can be rearranged, leaving alone this particular arrangement MATHEMATICS. If you are a JEE (IIT) aspirant I'd like to meet you. Mail me.
698365
What you are trying to do is actually part of a larger sorting algorithm called "Merge Sort".\n\nhere is an example:\nhttp://mathbits.com/MathBits/CompSci/Arrays/Merge.htm\n\nYou would keep merging the sorted arrays in set K into the "master" sorted array.  Thus, you would only be working with two sorted arrays at any one time.\n\nSearch the net for "Merge Sort" for other implementations.
724915
simple.  if you want to find the permutations of something of a set of numbers (or letters or objects), as long as you want to find just how many combinations there would be, you use factorial (!).  8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1...and that's your answer.
237350
Do you mean permutations?\n8P4\n= 8!/(8-4)!\n=8!/4!\n= 8*7*6*5\n= 1680
52716
I hope you can understand the pseudocode, it's pretty loose.\n\nConstruct two arrays P and Q.\nLet P be empty, and Q contain the n numbers you want to permute.\nTo start, call the function PrintPermute below passing in the arrays P and Q.\n\nPrintPermute( P, Q)\n{\n  if Q is empty, print the values in P in order and return\n  else\n\n  for each value q in Q,\n  {\n       Create the strings P' and Q' where P' = P + q and Q' = Q - q\n       Call PrintPermute(P', Q')\n }\nreturn\n}
1025267
Here is how:\n\n6+6+3\n6+5+4\n5+5+5\n\nnow change the first two around until you have all the possiblities.
784674
uhh,,i will have to ask my maths teacher in college,,\nthanx for the superb question,,
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!
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! :-)
367272
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
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
631878
If you are only concerned about the ways that you can place the 15 women between the 10 men, then you have 9 positions to place the 15 women.  (You don't have to worry about the arrangements for the men since there has to be one in between there is only one way to arrange.) These are the ways to do it (regardless of order).\n\na) 1 1 1 1 1 1 1 1 7\nb) 1 1 1 1 1 1 1 2 6\nc) 1 1 1 1 1 1 1 3 5\nd) 1 1 1 1 1 1 1 4 4\ne) 1 1 1 1 1 1 2 2 5\nf) 1 1 1 1 1 1 2 3 4\ng) 1 1 1 1 1 1 3 3 3\nh) 1 1 1 1 1 2 2 2 4\ni) 1 1 1 1 1 2 2 3 3\nj) 1 1 1 1 2 2 2 2 3\nk) 1 1 1 2 2 2 2 2 2\n\nNow to rearrange these, you would have to use the multinomial coefficient.  You have to use this because you have repeats: For example in a), you can switch two 1's around and get the same result, but moving the 7 will give you a different result.\n\nFor a), there are 9!/8!1! = 9 ways to arrange.\nFor b) and c), there are 9!/7!1!1! = 72 ways to arrange.\nFor d), 9!/7!2! = 36 ways.\nFor e), 9!/6!2!1! = 252 ways.\nFor f), 9!/6!1!1!1! = 504 ways.\nFor g), 9!/6!3! = 84 ways.\nFor h), 9!/5!3!1! = 504 ways.\nFor i), 9!/5!2!2! = 756 ways.\nFor j), 9!/4!4!1! = 630 ways.\nFor k), 9!/3!6! = 84 ways.\n\nSo in total, there are 9 + 72 + 72 + 36 + 252 + 504 + 504 + 84 + 756 + 630 + 84 = 3003 ways to place the 15 women between the 10 men.\n\nNow if after that you want to know how many ways there are to arrange if you do make a difference between men and between women, then multiply by the number of ways to arrange the men and women.\n\n3003*10!*15! = 14,250,102,106,034,995,200,000 ways.\n\nNotice this is less than the number of ways to arrange all 25 people regardless of gender.  That would have been 25! = 15,511,210,043,330,985,984,000,000.
1137758
1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
130493
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49\n\n6 f the above numbers
589231
yes here you can consider in terms of set like :\nHumanity={course1,course2,......}\nSociology=(course1,...........}\nScience=(course1,...........}\nMaths=(course1,...........}\nMusic=(course1,...........}\n\nhere n(Humanity)=3\nn(Sociology)=4\nn(Science)=5\nn(Maths)=6\nn(Music)=3\n\ntherefore no of ordered pairs will be equal to=3 x 4 x 5 x 6 x 3= 1080
198636
The number of combinations is given by\nnCr  =  n! / ( (n-r)! r! )\n\nThis formula works by calculating the number of possible permutations of r=5 choices from n=38 objects then dividing it by the number of ways the 5 objects can be arranged leaving just the number of possible combinations without worrying about their order.\n\nn!/(n-r)! = 38 * 37 * 36 * 35 * 34 = 60233040\n\nThis is the number of different possible permutations.\n\nr! = 5 * 4 * 3 * 2 * 1 = 120\n\nThis is the number of ways you can rearrange 5 objects. Each different possible combination of 5 from 38 objects can be arranged in 120 different permutations. Thus taking the total number of permutations and dividing by 120 will tell you the number of unique combinations.\n\n501942
1161271
In all these problems you are going to pick a set of objects or perform a set of actions.  \n\nIf the ordering of the outcomes makes a difference, then it's a permutation. If the ordering doesn't make a difference, then it's a combination.\n\nExample 1: The 10 runners run a race. They give medals for the first 3 places. How many different outcomes are there?\n\nThe order of finish makes a difference, since it determines who gets which medal, so it's a permutation problem.\n\nAnswer: 10x9x8 = 720\n\nExample 2: Three students are chosen from a class of 10 to clean after a food fight. How many ways can the group be chosen?\n\nIt doesn't matter who gets picked first, they all have to clean. Ordering doesn't make a difference, so it's a combination problem.\n\nAnswer: 10C3 = 10!/3!7! = 10x9x8/3x2x1 = 120.