Combination Question...?
968455
The ans is:\n\n27(26^4)\n\n26^4 + 26^5 = 26^4(1+26) => 27(26^4)
1096868
If I did the math right there are ten numbers 0,1,2,3,4,5,6,7,8,9\n\nThey must be put in a group of four. Thats 10 to the fourth power. So thats about ........  10,000 possible combonations. If your asking the possible combonation to a lock....... Well good luck unloking it.
342259
6! or 720 arrangements.\n\n2 * 4! or 48 with vowels at the ends, so it's a 1/15 chance.\n\nEdit: Yeah, I messed up.  Forgot the *2.
590394
In 36 possibles outcomes\nprobability of doubles\n1-1\n2-2\n3-3\n4-4\n5-5\n6-6\n6 in 36 or 1 in 6 or 16,666%\nOf the sum being 4\n1 +3\n2+2\n3+1\n3 in 36 or 1 in 12 or 8,333%\nIf it is combined probability you can sum the two\n16,66+8,33=25%
427494
The number of combinations is determined by multiplying the number of possible numbers of the first by the number of possible numbers of the second, third and last. If the any of the numbers could be used more than once, the answer would be 4 X 4 X 4 X 4 = 256. Since this problem says that the number can only be used once, then the number of possible choices for the first is 4, the second is only 3, the third is 2, and the last is only 1. So the way to arrive at 24 is 4 X 3 X 2 X 1. Maybe an example to make sure it is clear. If the first digit was 3, then the second could only be one of 5, 8, or 9. If the second was chosen as 5, then the third could only be one of 8 or 9. The last then could only be the one not selected as the third.
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/
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.
672664
first you figure out the number of possible outcomes you can have. which would be 6x6 = 36. \nThen you figure out how many combinations you can make to make it greater than 7:\n5 and 3\n5 and 4\n5 and 5\n4 and 4\n4 and 5\n3 and 5\nwhich is 6.\n6 over 36 is one sixth. therefore your answer is 1/6.
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
939247
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
695512
Answers choices are:\n     1, 4, 3, 2, 2,\n      3, 1, 4, 2, 1,\n      1, 1, 4, 2, 1\n      3, 2, 3, 4, 1\n      3, 4, 1, 2, 4
679588
Multiply it out...\n2y+6y+24x-24y-6x\nThen combine\n-16y+18x OR\n18x-16y
469878
10 * 9 * 8 * 7 * 6 = 30240 combinations
920768
One solution is\n0, 2, 4, 6, 8, 10, 12, 1, 3, 5, 7, 9, 11\nBut there is a huge variety of solutions\n\nSorry Chris I did not copy so here is another\n\n0, 11, 5, 2, 9, 3, 1, 4, 10, 7, 12, 6, 8
93645
d=5, g=1\n\nFirst I found all the combinations of 3 numbers (no repeats and no fours) that added to 13 (17-4) for b+c+d. Then I found all the combinations of 3 numbers (no repeats and no fours that added to 17 for g+h+i. Then I figured out which of these pairs of numbers had digits that didn't overlap, and had a pair of digits (d & g) that added to 17 minus the digits that, between them, were still missing (e & f).\n\nAnd, I'm a gal. :)
133515
I got it....easy solution is to list out all possible satisfying combinations and you come out with 21 out of 36 (7/12).\n\nThe OR in your problem is an inclusive-OR (meaning satisfying either condition satisfies the goal).  However, what I forgot to include was the subtraction where both outcomes occur.\n\n1/2 = just getting even on first roll\n1/6 = getting 1 on second roll\n3/36 = getting even on 1st and 1 on 2nd\n\n1/2 + 1/6 - 3/36 = 7/12\n\nThink of it as a Venn diagram.  If we do not subtract out 3/36, then we are double counting some results (namely 2-1, 4-1, 6-1).\n\n\n>>>>old post\n1/2 + 1/6 if it's an inclusive OR\nneed to get even or get the 1 on the die.\n\n\n1/2 * 5/6 + 1/2 * 1/6 if exclusive OR\n50-50 getting even but must have 2-6 on next die.  Then 50-50 on getting odd but must get the 1 on next die.\n\nOr maybe I'm just really wrong.
836747
2,3,1 (nothing goes there) ,6\n\nHope it helps :]