help me solve this Permutation / Combination problem please?
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!)
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
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.
18114
This is true only if the items are distinct. For example if the data set is the 26 letter alphabet and we take letters without replacement, then the number of permutations (order matters, so a,b is different from b,a) is 26*25 while the number of combinations (order does not matter, taking a, then b is the same as taking b, then a) is 26*25/2. \n\nIf our data set contains copies, like the letters in this sentence, then combinations of identical letters would not count doubly (so, a,a is not a seperate permutation from a,a).
868097
-\naside from 1,\n2,3,4,6,12
1157976
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
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
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."
57033
the answer is 3 because a-b|a-b & a-b|a so a-b|b
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
191858
If there are no other constraints, the roles can be assigned in \n15*14*13*12*11*10 ways. The first role can be filled in 15 ways and the second in 14 ways and so on. This is also called a permutation. See http://mathworld.wolfram.com/Permutation.html
182799
Sas will do it for you.\n\nHere is the article:\nhttp://support.sas.com/techsup/technote/ts498.html\n\nIf you know visual basic (and you can do this from excel using VBA) here is some code for generating combinations:\n\nhttp://www.codeguru.com/vb/gen/vb_misc/algorithms/article.php/c5607/
101541
This is kinda wierd without drawing it out, but I'll try to explain and you can come up with the picture.\n\n1)  Start Program\n2)  Accept Input\n3)  Add 1 to counter\n4)  If counter is not equal to 3; go to #2\n5)  If N2 > N3, do #6-8; if not, go to #9\n6)  Move N3 to a hold field\n7)  Move N2 to N3\n8)  Move the hold field to N2\n9)  if N1 > N2, do #6-8 with the N1 & N2 replacing each other\n10) if N2 > N3, do #6-8\n11) print them
900956
If you include combinations with no people and all people, the answer is 32 ways.  2^5.\n\n{ } = 1\n{A}, {B}, {C}, {D}, {E} = 5\n{AB}, {AC}, {AD}, {AE}, {BC}, {BD}, {BE}, {CD}, {CE}, {DE} = 10\n{ABC}, {ABD}, {ABE}, {ACD}, {ACE}, {ADE}, {BCD}, {BCE}, {BDE}, {CDE} = 10\n{ABCD}, {ABCE}, {ABDE}, {ACDE}, {BCDE} = 5\n{ABCDE} = 1\n\nThat's a total of 32 ways.\n\nAnother way to look at is to think that each person can be included or excluded.  There are two ways to include/exclude person A, two ways to include/exclude person B, etc.  That's 2^5 = 32.\n\nThis all assumes that ordering doesn't matter (that a combination of person A and B is the same as person B and A).\n\n(Note: if no people in the set is considered invalid, just subtract 1 and get 31).
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
331637
hahaha. do it yourself. ill start you off.\nfirst write out the 5 possibilities of the first letter. then for each letter, write out the 4 possibilities of the 2nd letter. then for each of those combinations, write out the 3 possibilities of the 3rd letter and for each of those write otu the 2 possibilities of the 4th letter and then theres only 1 possibility of the last letter for each of them.\n\nF L A S H\nF L A H S\nF L S A H\nF L S H A\nF L H A S\nF L H S A\nF A L S H\nF A L H S\nF A S L H\nF A S H L\nF A H L S\nF A H S L\nF S L A H\nF S L H A\nF S A L H\nF S A H L\nF S H L A\nF S H A L\nF H L A S\nF H L S A\nF H A L S\nF H A S L\nF H S L A\nF H S A L\n\nL F\n  A\n  S\n  H\n\nA F\n  L\n  S\n  H\n\nS F\n  L\n  A\n  H\n\nH F\n  L\n  A\n  S