combinations/permutations/algorithm/formula? please?
959412
The answer is 9! / (6!*3!).\n\nYou're dealing with combinations and not permutations, so we don't care which person is which... you really have nine chairs, six of which are occupied and three of which are empty. Forget about ABCDEF being the six people: Let X represent an occupied chair, and O an empty chair.\n\nIf you need to create a list of all possible combinations, just start with the X'es all to the left and the O's to the right, then do a bubble sort thingy.\n\nXXXXXXOOO\nXXXXXOXOO\nXXXXXOOXO\nXXXXXOOOX\nXXXXOXXOO\nXXXXOXOXO\nXXXXOXOOX\nXXXXOOXXO\nXXXXOOXOX\nXXXXOOOXX\netc.\n\nIf you grind your way through the whole list, you'll find there are 84 combinations. There's some neat stuff going on here with triangular numbers: you can see that there's one arrangement where the first O is in the 7th position, three where the first O is in the 6th position, six where the first O is in the 5th position... follow the pattern and you'll see that there are 1+3+6+10+15+21+28 combinations in the list. That adds up to 84.\n\nHopefully that helps!
842276
1,2,3,4,6,8,12,16,24,48 \n\nlolz hope this helps :)
262
Permutation\nSuppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.\n\nBecause order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:\n\n      CA   CT   AC   AT   TC   TA\n\nCombination\nWhen we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:\n\n      CA   CT   AT
747450
Sounds like Combinatorics.  Combinations, permutations, a bit of graph theory, discrete math, I usually describe it as "Advanced Counting".
35955
like this???            \n              \n                x x x x x \n                x x x x x \n                x x x x x \n                x x x x x \n               (4x5/5x4)
982580
Use binomial coefficients to solve this.\n\n| n |           n!\n|    |   =  ----------\n| k |        (n-k)!k!\n\nn is the number of possibilities and k is the number of unordered outcomes.\n\nCombination of six numbers from ten possibilities.\n\n| 10 |       10!          3628800         \n|      | = ------------ = --------------- = 210\n|   6 |    (10-6)!6!     (24)(720)\n\nCombination of four numbers from ten possibilities.\n\n| 10 |\n|      |\n|   4 |\n\nYou should be able to figure out the second one on your own.\nHere's a link on factorials, just in case you don't know what they are.\n\nhttp://mathworld.wolfram.com/Factorial.html
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
378327
I don't think there's space here for me to do it how you,  but i'll get you started.\n\nFirst do all the combinations that start with F\nFirst, FL\nFLASH\nFLAHS\nFLSAH\nFLSHA\nFLHAS\nFLHSA\n\nDo it sysematically. FLA then FLS then FLH\nNow FA\nFALSH\nFALHS\nFASHL\nFASLH\nFAHLS\nFAHSL\n\nAnd just keep doing this, working systematically through every letter. \n\nHope this helps.
88776
Are you looking for a set of 8 consecutive whole numbers that when you add 7 of them together in different combinations you receive those 3-digit responses?\n\nWell you've only listed 6 possibilities and only one of those is odd.\n\nStarting at 29 the lowest combination is 224 and the highest is 231.  Starting at 28 the lowest is 217.
230780
1,2,3,4,5,6  Try those!  (Same odds as any other combination of numbers!)
236101
permutation?\n\ni'm not sure what you mean by "see how many it is."
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
1103566
# combinations of 3 elements out of 5 = # combinations of 2 elements out of 5 = 5*4/2 = 10\n\n5 ways to pick one zero\n4 ways to pick another zero\n2=2! permutations per pair of zeros.
242452
This is a typical combination problem.\n\nformula for combination is n!/r!(n-r)!\n\n= 9! / 5!(9-5)!\n= 9! / 5!4!\n= 126 (ANS.)\n\nfactorial ex. 5! = 5 x 4 x 3 x 2 x 1\n\nDONT FORGET TO RATE THIS AS BEST ANSWER
16141
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
833994
You might have seen this formula\n  n!/p!(n-p)! . What it means is : total permutations divided\nby permutations per combination.  The total number of\npermutations of 4 numbers  within 10,000 numbers(0000-9999)\nis  10000!/(10000-4)!=\n(10000)(9999)(9998)(9997).\nThe number of permutations per combination of 4 is 4!.\nSo the answer is \n(10000)(9999)(9998)(9997)/\n{(4)(3)(2)}.\nThat's (1250)(3333)(9998)(9997)= \n416,416,712,497,500
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! ]