1-2-3-4-5-6 List total 4 number combinations?
406750
6*5*4*3 = 360 if order matters.\n\n6*5*4*3 / 4! = 15 if order does not matter
340631
First you must consider that the first digit can not be zero, otherwise it would not be a 6 digit number.\nThen you must consider how you can arrange odd or even within the digits. (O=ODD, E-EVEN)\n1.OOOEEE\n2.OOEOEE\n3.OOEEOE\n4.OOEEEO\n5.OEOOEE\n6.OEOEOE\n7.OEOEEO\n8.OEEOOE\n9.OEEOEO\n10.OEEEOO\n11.EOOOEE\n12.EOOEOE\n13.EOOEEO\n14.EOEOOE\n15.EOEOEO\n16.EOEEOO\n17.EEOOOE\n18.EEOOEO\n19.EEOEOO\n20.EEEOOO\nFor each of the #1-10 above we can consider 5 possible digits for the odds and 5 for the evens\nso for each there are 5*5*5*5*5*5=15625 different digit combinations.\nFor #11-20 we must consider that the first digit can not be zero, so there are 4*5*5*5*5*5=12500 different digit combinations for each.\nSo there is a total of 10*15625 + 10*12500=281250 different numbers with 3 odd and 3 even digits.
532000
How about 9! 51, 52, 55, 56, 57, 58, 59, 60, 61
237350
Do you mean permutations?\n8P4\n= 8!/(8-4)!\n=8!/4!\n= 8*7*6*5\n= 1680
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
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
178173
Simply put the numbers 1 through 4 on the die, repeating two of the numbers. Is there a further requirement that you haven't given ?
748483
The answer is 119\n\nTake 2 at a time end up with 1 (59 times=118 one left)\nTake 3 at a time end up with 2 (39 times=117 two left)\nTake 4 at a time end up with 3 (29times =116 three left)\nTake 5 at a time end up with 4 (23times=115 four left) \nTake 6 at a time end up with 5 (19times=114 five left)\nTake 7 at a time end up with 0 (17times=119 o left)\n\nI am dead sure this time = ]\n\nThe trick is look at 5 at a time. there is 4 left this means the last digit must be 9. So look at all possible answer of 7s that ends with 9 and work it out from that!!!!!!!!!!
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,....
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
815499
The correct answer is 15, but the proper formula is (6C4)/2\n\nBy the way, you can check this by figuring out how many groups of 2 you can have, since every group of four being chosen is the same as a group of two not being chosen.  For the first person, there are six people to choose from.  For the second, there are five.  However, since picking Person I and Person II is the same as picking Person II and Person I, you have only half as many possible combinations, or 15.
906203
Assuming you cannot reuse a digit unless there is more than one copy in the list...\n\nYou could do it this way...\n5 one-digit numbers: 2, 5, 6, 7, 9\n22 two-digit numbers: 22, 25, 26, 27, 29, 52, 56, 57, 59, 62, 65, 66, 67, 69, 72, 75, 76, 79, 92, 95, 96, 97\netc.\n\nBut it would take a while. The answer is: at least a few.
1186083
All four hands can have 6-5-2 arrangement.\n\n6520\n0256\n5602\n2065
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
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.
952696
n + (n - 5) + (n + 6) \n3n + 1
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.