permutation or combination problem?
363153
15! gives you the answer to how many combinations of placings you can get for all 15 teams.  That's not what you want.  You're just looking at the top three.  The answer is 15x14x13. Why? There are 15 possible teams that could get 1st place. Pick one. That leaves 14 possible teams that could get 2nd. Pick one. That leaves 13 possible teams that could get third. The number of possible combinations of this is then equal to 15x14x13.
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.
381290
I have an iterative method that is good for at least 80% of these types of problems.\n\nFirst, write down the series.\n\nNext, above the line you just wrote, write the difference between each pair of elements. In this case, it will be 1, 2, 4, 8, .... In this case, the answer is obvious - they are powers of two. The next one in the differences line should be 16, so the next on the original line should be 15+16=31. This will allow you to get any member in the sequence.\n\nIf the first order of differences doesn't give rise to a discernable pattern, try the differences of differences and so on.
633977
Permutation = in how many ways can you arrange a group of objects. For example the permutations of a, b, c are:\na-b-c, a-c-b, b-a-c, b-c-a, c-a-b, c-b-a\n\nCombination = in how many ways can you select a certain un-ordered sub-group from a larger group (e.g. select 3 students from a group of 31). For example, combination of a, b, c in groups of 2:\na&b, a&c, b&c
775189
the following is the total permutation of\nthe four numbers:-\n\n1234\n1243\n1324\n1342\n1423\n1432\n2134\n2143\n2314\n2341\n2413\n2431\n3124\n3142\n3214\n3241\n3412\n3421\n4123\n4132\n4231\n4213\n4321\n4312\n\ntotal amount of permutations=4!\n4*3*2*1=24\n\nnotice that 4 is grouped with 123 six times\nnotice that 3 is grouped with 124 six times\nnotice that 2 is grouped with 134 six times\nnotice that 1 is grouped with 234 six times\n\nthe six permutations of the three other\nnumbers =3!=3*2*1=6\n\nthere are two permutations of the other\ntwo numbers =2!=1*2\n\nhence,total for four numbers= 4*3*2*1\n=24\n\nthey are symmetric since each number\nhas to appear the same number of\ntimes in the permutation\n\ni hope that this helps
89444
The answer is "i" because there is no number, that when mutliplied by itself, will give you -1.  Think about it: the only possible combination that would make sense is 1 X 1, which equals 1...not -1.
527818
The Fundamental Counting Principle gives us 12 combinations in this case, so if the first 12 are wearing different things, then the 13th must be wearing one of the other 12's combos. The answer to your second question is 25.
215950
almost ... its 1000 to 9999 \n\ni.e 9000 combinations :)\n\nAnother way to do this ... \n\nexcept the first digit all can have a 0 .. so the possible combinations are : \n\n9*10*10*10 = 9000\n\n( 9 possible digits can be put in the first and 10 in the 2 thru 4)
660290
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, 50, 51, 52, 53, 54, 55, 56.\n\nThe winning combo is amongst the nice hot lucky winning numbers I listed above. Good luck!!!
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.
469878
10 * 9 * 8 * 7 * 6 = 30240 combinations
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.
1029989
The answer is 120 "lock combos" if I've understood your question.\n\nYou can't use a number twice, but different orders are allowed, but a combo must end in 0. So, for example, 2450 and 5240 are valid combo numbers. If that's the case, the answer is 120.\n\nYou must choose 3 distinct numbers in any order from the numbers 2 4 5 6 7 8. You then add a 0 on the end of these 3 to get a  valid 4 digit combo number. The number of ways of doing this is simply 6x5x4 = 120. Note, if digits could be used more than once it would 6x6x6 = 216.\nAs far as using formulas go, you would use the permutation formula: \n\nnPr = n!/(n-r)! In your case n= 6 and r = 3\n\n6P3 = 6x5x4 = 120
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).
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.
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.
1180538
Nope, that is the combination of {x,y,z} taking 2 objects at a time, because combination has no regard for specific order. The permutations for {x,y,z} taking two  at a time are: xy, xz, yx, yz, zx, zy.