ken made 18 pnts. in 1 b-bal game how many possible combinations r there of 2pt. shots and 3pt. shots r there?
144314
He could have 9 2s, 6 2s and 2 3s, 3 2s and 4 3s, or just 6 3s. That is 4 different combos.
1144298
Have a play in game between the 8 and 9 seeds.\n\nThen:\nRound one\n1 v. 8/9\n2 v. 7\n3 v. 6\n4 v. 5\n\nRound two:\n<1/8/9> v. <4/5>\n<2/7> v. <3/6>
192695
You have the following lists:\nB B B\nB B G\nB G B\nG B B\nB G G\nG B G\nG G B\nG G G\n\nSo you have 8 variations and of those 3 are 2 boys 1 girl so the probability is 3/8.
15239
The correct answer is 190. It would involve any spare (all pins down on two balls) in the first frame, then single-pin spare conversions in frames 2-9 and a 9-count, spare and another 9-count in the tenth and final frame.
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
907310
5\n\n1s,2s,2px,2py,2pz
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.
852237
I know one for the doubles.\n\nIt's the doubles baby, Let's go. Let's go.\nIt's the doubles baby. Let's start with zero.\n\n0 x 0 = 0      Oh!\n1 x 1 = 1      Fun!\n2 x 2 = 4      More!\n3 x 3 = 9     That's fine!\n4 x 4 = 16   That's keen!\n5 x 5 = 25   We're Alive!\n6 x 6 = 36   That's were we get our kicks!\n7 x 7 = 49    Go 49er's\n8 x 8 = 64    Knock on the door!\n9 x 9 = 81    Isn't this fun!
325534
One player shoots the ball from any spot on the court.  If he makes it, everybody else must make the same shot.  If they all make it, nothing happens, and the process starts over again with the first person taking a new shot.  If any player misses, he or she gets the letter H, and the next player takes a new shot.  The first player to get HORSE loses.  If you want to get more complex, the first shooter can call "Glass" or "Swish" before his shot, in which case he and all subsequent players must bank or swish the shot, or else it counts as a miss.  The game is sort of hard to explain in words though, and your best bet is to find someone who knows the rules and actually learn while playing.  If this is not possible, I hope my directions help.
894091
None. Two odds make an even
1169447
The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
267207
the options given are one and two player
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
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!!!!!!!!!!
724427
1024 matches. This should be on the mathematic section. Everyone know that an elimination match would always require <number of player> - 1 to finish. \n\nThe simple logic is actually: We're gonna create 1024 losers, how could we create 1024 losers with less than 1024 match? Can't have 1024 losers with only several matches.\n\nEach player would play the match 10 times (2^10 = 1024), and exactly one player would play 11 matches. (to complete up the 1025 players)
242656
This one is easy-assume a 50% chance each time, so its 1/2 X 1/2 X 1/2 X 1/2, so 1 chance in 16.  There is another way of figuring this, using combinations, like sn777 did, but sn777 made a logical error.  There are these possible combinations: 4 m, 3m 1F, 2m, 2F, 1m 3F, and 4F. So just looking at that, you would think its 25%. BUT, there is only one combination of 4M, while there are 4 combinations that make 3m 1F, and 6 combinations that make 2m and 2F, etc-so the total possible combinations of 4 children is 16, and 4M is only one of those.
185383
There are a total of 63 games to be played (every team will lose exactly once, except the entire winner of the tournament).\n\nIn each game, there are exactly two possible outcomes (given the results of the games leading up to it).\n\nTherefore the number of completely-filled brackets is 2^63, that is 2 to the 63rd power, that is 2*2*2*...*2 (63 times).\n\nThis is an *enormous* number.  To give you an idea, it is 19 decimal digits long.\n\nIn general, if there are n teams, then there are 2^(n-1) ways to fill out the bracket completely.  This formula matches your examply (although, with 2 teams there are actually 2 possibilities).