for mn= 20! and 0<m/n<1, how many ordered pairs (m,n) are there?
929018
mn=20!\nmn= 20*19*18*17*......*2=20!\nif m=20; n=19!\nif m=20*19; n=18!\nand so on.\nso m=20!/n; n=i! .... i 20 to 1\n\n0<m/n<1 means that m Not eq 0; and m < n\nby solving the above,\nm will remain < n only till iteration 8\n\nSo the ordered pairs are:\nm=20!/20!; n= 20!\nm=20!/19!; n=19!\nm=20!/18!; n=18!\nm=20!/17!; n=17!\nm=20!/16!; n=16!\nm=20!/15!; n=15!\nm=20!/14!; n=14!\nm=20!/13!; n=13!
907310
5\n\n1s,2s,2px,2py,2pz
147821
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines
376218
Not sure how many you need, but here are the first few:\n2,3,5,7,11,13,17,19,23,29,31, 37,41,43,47,53,59,61,67, 71,73,79,83,89,97,101
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
868097
-\naside from 1,\n2,3,4,6,12
230538
0!=1\n2!=2\n4!=24\n6!=720\n8!=40320\n10!=3628800
517585
Starting with 0, the sequence is the square of that number:\n\nf(n) = n^2. \n\nf(0) = 0 (0^2)\nf(1) = 1 (1^2)\nf(2) = 4 (2^2)\nf(3) = 9 (3^2)\nf(4) = 16 (4^2)\nf(5) = 25 (5^2)\netc.\n\nNote, it isn't clear if your indexes are going 0,1,2,3,4,... or are going in descending order 0,-1,-2,-3,-4... the resulting sequence would be the same.\n\nf(0) = 0 (0^2)\nf(-1) = 1 (-1^2)\nf(-2) = 4 (-2^2)\nf(-3) = 9 (-3^2)\nf(-4) = 16 (-4^2)\nf(-5) = 25 (-5^2)\netc.
264022
This is the combination of 40 things taken 10 at a time :\n40 C 10 = 40! / (10!30!)= 847,660,528\n \nCombinations do not consider order or arrangement (i.e. a,b as against b,a).  When order is considered we take the permutation instead.
197475
this one is pretty easy\nafter 46 i will give u the next no.\n\n54,55,62,63,69,70,75,76\n\nc it goes like that 80,81,84,85,....
49819
Hold on....10....20......25.....29..30....25..29\n\n25
1008114
This series is n(n+1)\n\n2,6,12,20,30,42,56,72,90,. . .
727389
1) 21\n2) 52\n3) 40\n4) 80\n5) 104\nIn each pattern you are adding a number to the one before it.\nEx: 1 + 0 = 1, 1 + 1= 2, 2 + 1 = 3, 3 + 2 =5, 5 + 3 = 8, etc.
836747
2,3,1 (nothing goes there) ,6\n\nHope it helps :]
218117
Correct that n! = n x (n-1) x (n-2) x ...\n\nFor 0! consider this:\n\n4! = 4 x 3 x 2 x 1\n\n3! = 3 x 2 x 1\n\n2! = 2 x 1\n\nFrom this we can see that n! = (n+1)! / (n+1)\n\nTherefore, continuing with decreasing values,\n\n1! = 2! / 2  =  (2 x 1)/ 2 = 1\n\nand it follows that\n\n0! = 1! / 1  =  1 / 1  = 1\n\nEstablished for 0! without exclusive definition.
432827
(n-1)!/(n+1)! = (n-1)!/((n+1)n(n-1)!)\n= 1/((n+1)n)\n= 1/(n^2+n)
713219
1 isn't a multiple of 4, but that was just a typo, right? :^)\n\nStrictly speaking, if the question asks "..for all multiples of 4 between 1 and 20", the answer would be "4,8,12,16" because "between" mathematically does not include "equal", unless the question had been written "..between 1 and 20, INCLUSIVE", which indicates that the endpoints would be includable.  Then you would add "20" to the list (but not 1 ; 4 times what whole number gives you 1?)