1-2-3-4-5-6-7 List total 4 number combinations?
452386
7*6*5*4 = 840 if order matters.\n\n7*6*5*4 / 4! = 35 if order does not matter
757339
one combination of 5 digits and 120 permutation.\n\none combination 5 digits can be picked in only one way \n\n19367 91367 all are same.\n\nfor permutation\nboth the above are different\n\n1st letter 5 choices , 2nd letter 4 choces so on .\n\nnumber 5*4*3*2*1 = 120 choices
845925
If you problem, or part of it, is:\n\nHow many 8 digit numbers can be formed from this list of digits?\n8! / (3! * 2!)\nThe number combinations of digits has tobe reduced by 2! because of the 2 "6"s.\nThe number combinations of digits has tobe reduced by 3! because of the 3 "2"s\n.\nBut since the number cannot exceed 65,500; all larger numbers must be eliminated. Subtract the number of 8 digit numbers that can be produced from the list that start with "9". Likewise "7". Likewise "69", "67", "66", "659", ""657", and "656".\n\n\n\n(Try to state your problem clearly and completely.)
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.
827303
a(n) = a(n-1) + sum of digits of a(n-1)  \n \n25, 32, 37, 47, 58, 71, 79\n\nNext is 79 + 16 = 95.
230780
1,2,3,4,5,6  Try those!  (Same odds as any other combination of numbers!)
724915
simple.  if you want to find the permutations of something of a set of numbers (or letters or objects), as long as you want to find just how many combinations there would be, you use factorial (!).  8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1...and that's your answer.
469878
10 * 9 * 8 * 7 * 6 = 30240 combinations
786009
-2(x-3) + 4(x+7)\n-2x+6   +4x+28\n2x+34
1118957
The first number has to be (x+6) and the second number is (x).  There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93).  60 plus 8 (68) does not equal 06, or 6.  71 plus 8 does not equal 17.  82 plus 8 does not equal 28.  93 plus 8 does not equal 39.\nUnfortunately, this problem is impossible.\nSorry I couldn't be of more help.
868097
-\naside from 1,\n2,3,4,6,12
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
819558
Theres 7 choices for the first digit, 6 more for the second .. 2 for the sixth. So 7*6*5*4*3*2 = 5040.
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!!!
1132286
#4, #2, #4, #4, #2, #4 #1...repeat......
526416
add the numbers\n15\n19\n42\n19\nnext number equals 15.....     and each has a 2 and a 7 in it\n\ntry 2742 sounds like a good answer for neopets lenny conundrum answer possibility but not fully sure..... lucidlyinsane
1060907
-1, +4\n12, 11, 15, 14\n*18, 17, etc.