What is combinatorial approach?
455707
there's are so many different approach's....\n\ncitebase.eprints.org/cgi-bin/citations?id=oai:arXiv.org:nlin/0511015 \n\nhere's one site i'm sure you could link from there....
805688
U can use either Kruskal's algorithm or prims algorithm. More information about these algorithm u can find on internet.\n\nPrims algorithm is based on adjacency matrix.
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
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
893652
with 1q, 2 ways (0,1 n)\nwith 3d, 1 ways (0 n)\nwith 2d, 3 ways (0,1,2 n)\nwith 1d, 5 ways (0,1,2,3,4 n)\nwith 0d, 7 ways (0,1,2,3,4,5,6 n)\n\nTotal of 18 ways
636772
k=1/4 or k=2 or many more possibilities
260821
labeling approach\nsymbolic interactionism\nrational choice\nrole theory
92016
When dealing with the occurrence of more than one event, it is important to be able to quickly determine how many possible outcomes exist.\n\nFor example, if ice cream sundaes come in 5 flavors with 4 possible toppings, how many different sundaes can be made with one flavor of ice cream and one topping?  \n\nRather than list the entire sample space with all possible combinations of ice cream and toppings, we may simply multiply 5 • 4 = 20 possible sundaes.  This simple multiplication  process is known as the Counting Principle.
419391
To find out the number of combinations taken 4 at a time (and assuming without replacement or that a number cannot be used more than once), we use the formula\n\nnCr where n is the numbers of items and r is how many are taken at a time. in this case, n = 10 and r = 4\n\nnCr = (n!)/[(n-r)!r!]\n\n= (10!)\n--------------\n(10-4)!4!\n\n= (10)(9)(8)(7)(6)(5)(4)(3)(2)(1)\n--------------------------------\n(6)(5)(4)(3)(2)(1) (4)(3)(2)(1)\n\nsimplifying we get\n\n= (10)(9)(8)(7)\n----------------\n  (4)(3)(2)(1)\n\nsimplifying again\n\n= (5)(3)(2)(7)\n\n= 210 combinations
798663
Francopit was by far the closest yet.  Since there are only 2 kids, an exhaustive solution really makes sense.  With 2 kids, there are 4 possibilities of male/female combinations.  Listing the older one first they are:\n\nMM\nMF\nFM\nFF\n\nSince we know there is at least one male, the problem reduces to three possibilities:\n\nMM\nMF\nFM\n\nNow for each case there are two choices depending on which one is the king/queen.  Using caps for the king/queen:\n\nMm\nmM\nMf\nmF\nFm\nfM\n\nThe problem list a king not a queen so all the 'F' cases are discarded.  Francopit said, mM is not possible since the oldest male is king.  This leaves:\n\nMm\nMf\nfM\n\nBased on the problem statement, these are all the possible configurations.  In one, the king has a brother, in the other 2 he has a sister so the probability of a sister is 2/3.
44177
I'm sorry, but I don't think that you have actually asked a question here, or perhaps you have left off some very pertinent information. Unless you just wanted the above statement to be explained.\n\nIn this instance, the universe of discourse just means all the points contained in the set S = (p(x,y)), which is all the possible combinations of x = 1,2,3 and y = 1,2,3. For example, (1,3) and (2,2) and (3,1) would all be points contained in the universe of discourse for p(x,y).
725814
This is a very difficult counting problem.  I will show one possible grouping and count the number of ways it can be done.\n\nTo put just 1 boy in group 1, there are 10 ways to pick the boy and "combination of 15 things taken 4 at a time" ways to pick the girls.  This is 10 * (15*14*13*12 / 1*2*3*4) = 10*1365 = 13650 ways to pick 4 girls and 1 boy.\n-----------------\nThere are many grouping scenarios; let us start with 1 boy, 1 boy, 1 boy, 2 boys, 5 boys.\n\nWe know there are 13650 ways to pick a single boy for group 1.\n\nTo then pick a single boy for group 2 from the remaining children, there are 9*(11*10*9*8 / 1*2*3*4) = 9 * 330 = 2970 ways.\n\nIf a single boy is then selected for group 3, there are 8 * (7*6*5*4 / 1*2*3*4) = 8*35 = 280 ways.\n\nGroup 4 must have at least 2 boys.  There are 7 boys left, and we must choose 2 of them.  So there are 7*6/1*2 = 21 ways to choose those boys with the remaining girls.\n\nThe last group would have 5 boys.\n\nSo just for this one possible way of grouping - 1 boy, 1 boy, 1 boy, 2 boys, 5 boys, there are \n13650 * 2970 * 280 * 21 * 1 = 238,378,140,000 ways to arrange them.\n\nTHIS MAY NOT BE THE CORRECT FIGURE FOR THIS GROUPING SCENARIO - THERE MAY BE DUPLICATES, IN WHICH CASE THIS FIGURE IS TOO HIGH.  I did not know how to remove them from the figure.  The 13650 is correct for the first, but then we need to look at how many ways are there to grab 4&1, 4&1, 3&2, 0&5 from the remaining children.\n\n\nWe must also account for other grouping scenarios, such as 2, 2, 2, 2, 2 boys; 1,1,2,2,4 boys; 1,1,1,3,4 boys; 1,2,2,2,3 boys, and some others.  For each of these grouping possibilities, the above such calculations would have to be done.  The sum of all the grouping possibilities is the total.\n\nSo this is a very difficult counting problem.  I would think you should search for more tools from your professor/teacher, or book.  There may be some more advanced way of saving time rather than go through each grouping scenario one at a time; though that will give the correct answer.
294686
The number of permutations of a string of n unique objects is n!. \nThis can be derived by the fact that if you have the set of all permutations of that string, all the elements of the set can be generated by first choosing an object to be first in the string, then a second object from the n-1 remaining objects to be second in the string, and so forth. By the fundamental principle of counting, one can thus generate n*(n-1)*(n-2)*...*1 = n! elements.\nIf some of the objects in the string are not unique, ie., the number of permutations of the string (E, E, E, V, R, S, T), we can show that the number of unique permutations is the same as the number of permutations given by pretending the objects are unique divided by the number of permutations of the non-unique objects.\nThis can be shown as follows: Suppose P(n) is the number of permutations of n unique objects. Suppose that P(n, r) is the number of permutations of n unique objects and r non-unique objects. Then by the fundamental principle of counting, if we replaced each non-unique object in the string from a set of r objects, we have P(n, r)*P(r) = P(n) permutations. Thus P(n, r) = P(n)/P(r) = n!/r!.
803007
I don't understand your question.  Do you mean using the associative law?\n\nabc*de = \ncba*de =\nbca*de =\nacb*de =\ncab*de = \nbac*de \nNow all of these are multiplied by the two ways to postiion de\n\nThat's 3! = 6\ntimes  2! = 2\n\nfor a total of 12.  \n\nFor a 5th grader I'm sure that you're just supposed to list all 12.  For convenience, note that I've put each letter in the middle and switched the other two.  For neatness, I'd recommende just putting another column to the right of each of these 6 with de and ed .\nWhen doing counting like this, called combinotorics, it's helpful to arrange things neatly.
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!)
601571
You meant a combination of qualitative and quantitative methods. That's good way to proceed especially if it's a new research area.
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.