If 6 spikers, 4 setters and 4 spotters are looking for placement. How many different teams can be assembled?
969926
Spikers. You choose 4 out of 6 of them, meaning 2 are left out. \nChoose the first to be left out: 6\nChoose the 2nd to be left out from the 5 remaining. 5\n30 possibilites. But choosing person A then B, is the same as B then A, so that means 30 / 2= 15 different ways of choosing spikers\n\nSetters: Once again choose 2 of them. First:4\n2nd from the 3 remaining: 3\nDivide by 2 again: 4*3 /2 = 6 different ways of choosing setters\n\nSpotters: We must take all of them. 1 way\n\n15*6 = 90 different teams that can be assembled.
1081375
It's a random selection but with guidelines:\n\nThere cannot be more than 2 European teams in one group, and 1 team per continent in each group.\nHighly ranked (England, Argentina, France, Mexico, Italy), home teams (Germany)  and defending champions (Brazil) are placed at the top of each group
692322
there are sixx\n\nif ur runninf a 4-2\n\nleft back is a passer, middle back is a passer, and right back is a server.\nright front is a hitter, middle front and left front are both hitters\n\nif ur running a six-2 and not a 4-2, and you have first serve,\n\nright back is a server, then runs up and becomes a setter. middle back is a passer, and left back is a passer.\n\nright front is a 5 hitter, or will back up and become a passer. middle front is a \nhitter along with right front.\n\nhope that clears it all.
852587
well there are 6 players on the court from one team at a time. 7 if you have a DS. it's not a timed sport you play for points. you play by getting the ball over the net by passing or setting or hitting without hitting it out of bounds. you score by having the other team hit it out or having them not get it over in their three tries. you win by winning 2 out of three games to 25 or 3 out of 5 fames to 25. depends on the rules. basically the back row wants to pass the ball to the setter so the setter can set up the hitter to slam the ball in the other teams face =]
246896
Close, but no cigar...\n\nABCDE FGHIJ KLMNO\nCJDMG OEKHB INALF\nGFBKL NAOJC DHIEM\nEKNCI LMGAH JOFBD\n\nConstraint: Fifteen riders yields (15 * 14 / 2) = 105 unordered pairs.  A single five rider heat yields (5 * 4 / 2) = 10 unordered pairs.  Over twelve heats that makes 120 unordered pairs -- a little bit more than the 105 that we want to cover.  So we want to have each pair of riders be together in at least one heat, with no pair of riders being together in more than two heats.\n\nResult: Not quite -- 21 "head-to-head" matchups do not take place.  But no pair of riders are together more than twice.\n\nAK BI BM BN CF CH CL\nDK DL DN EF EG EJ EL\nFM GN GO HN IO JK JL\n\nConstraint: There are four sets and five positions.  So we want to have each rider in a different position in each heat.\n\nResult: Constraint is met!\n\nConstraint: There are three heats per set and fours sets. So we want to have each rider be the first seat of a set, the second heat of a set, and the third heat of a set (obviously the fourth set will double up on one of the heats).\n\nResult: Not quite -- 5 riders miss one of the first/second/third heats in a set.\n\nC:310 D:202 G:220 H:031 O:022\n\nAlternative:\n\nABCDE FGHIJ KLMNO\nJDLGM HOEBK INFCA\nLFOAH DEGKN MIBJC\nEHNMI OCJFD GKALB\n\nThe same on the first two constraints.  On the third constraint there are now 7 riders (up from 5) missing one of the first/second/third heats in a set.  But they are all 2-2-0 with none of the 3-1-0's of the first schedule.\n\nAJ AM BF BN CG CH CK\nCL DH DI EF EJ EL FK\nFM GO IK IL IO JK JN\n\nA:202 D:220 E:220 H:220 K:022 L:202 M:202
198261
The least number would be 6 but in case someone is sick or can't play you might want to have substitutes. Also if you have enough for two teams that's nice because in practice you can see how you play against another team. This can help you play better. But really all you need is 6 people.
925525
13 plyrs/tm is probably too many. you wouldnt get enough teams to have a decent league, unless your school is massive or soccer-crazed. i would suggest play 5 v 5 or 6 v 6 games. this way each team would only require 6-8 players and you'd probably get mor e interest.
801178
Depends. There are usually 6 players on the court.\nMy school team had about 12.
594124
twenty is a great number.  the more people you have the cooler looking pyramids and formations you can have. larger squads look very impressive at competitions.
619739
Basics: 15 players each side.\n2 periods of 40 min.\npass ball backward\nTackle only the ball carrier\nTo score a try (5 pts) put down the ball at the ingoal of the oponent side. his give you right for a goal kick (2 more pts)\npenalty kick you can go for goal kick aswel (3 pts)\n\nIf you want detail go to http://www.irb.com/Laws/
121477
From your question I am not totally clear on what you need, but a classic 6 round tounement will start with 64 teams or players.\n\nrd 1 - 64 teams\nrd 2 - 32 teams\nrd 3 - 16 teams\nrd 4 -  8 teams (quarters)\nrd 5 - 4 teams (semis)\nrd 6 - 2 teams (finals) \n\nIf you have only 57 teams in your golf tourney, then I would Rank and list all the paired teams from 1 to 57 (or blind draw into spots 1 to 57). Then make up 7 teams to bring the first round up to 64 teams. Number them from 58 - 64 and call them Byes. \n\nThen in Round 1:\nTeam 1 plays team 64 (bye), team 2 plays team 63 (bye) and so on until Team 32 plays Team 33.\n\nRound 2: You have a standard 32 teams left. Team 1 plays winner of Team 32 vs 33, Team 2 plays winner of team 31 vs team 34 and so on...
1161085
There are 4 teams in each group to start off.  They all play one another.  Top two teams move on.  Then, the remaining teams play in a bracket-style tournament, which is single-elimination.
934917
C is at the end, so its just the other 8 positions to play with.\n\nCount how many permutations of those 8 there are.\n\nThen count how many there are where A and B stand side by side.\n\nThen subtract those out.\n\nEdit:  Poster below me is way off.  without any restrictions, 9 men could be arranged in over 360000 ways.   Not 81.\n\nBasically, you can take 'C' out of it.  There are 8 spots left, and 8 men.  Spot 1 can hold any of the 8.  spot 2 any of the remaining 7 and so on, giving 8X7X6X5X4X3X2 which is 8! possibilities.\n\nThat's 40320.\n\nOf those 40320, some involve A and B being side by side.  How many?  Well, they could be in slot 1&2, 2& 3 .... up to 7&8.  And they could be A-B or B-A.  That's 14 possibles.  For each of those 14, the other 6 spots could be filled 6! ways.\n\nSo there are 14 X 6! = 14 X 720 = 10800 combos with C in the 9 spot and A & B side by side.\n\nTherefor, of the 40320 with C in the 9 spot, 40320-10800 = 30240 possible arrangements have C at the end, but A and B NOT side by side.
964232
there are 22 players per side 18 on the ground 4 on the bench(used in rotation{interchange}) 3 in the middle, six in the back line 6 in the forward line 3 on ballers which includes all three zones(full back line, half back line, center line, half forward line, full forward line, on ballers and the interchange bench) the aim of the game is to kick the ball through the two middle posts which = 6 points either side of them = 1 point \n\nthe ways you can dispose of the ball is either to kick it or handball it to one of your  team mates it is a contact sport and given the speed these players travel at will put someone like us on our bottoms it is more dangours more then all other contact sports in australia for example leauge and union \n\nfor more information see http://afl.com.au/
178726
volleyball - 6, basketball - 5, handball - 1 \nthese are the minimum number of players to play for 1 team at 1 time. not including substitutes.
1065918
To get your answers, imagine what has to happen.  \n\na) First, you have to pick 1 of the 4 white balls out of the 12 balls there.\n\n2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.\n\nSo, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.\n\nb) First, how many ways can 5 people line up?  5!.  How many ways are correct (assuming they all have different names, only 1).  So, the prob is 1/5! or 1/120.
611582
I believe that it goes like this:\n\nSuppose we have 26 jumpers, named A, B, C, D, etc jumping in alphabetical order.\n\nA, B, C, D, and E take their first jump. Suppose they all miss. They take their second jump in the same order. If, say B and D clear the height, then F and G are added to rotation.\n\nI can't find a reference, but I believe that F and G take their first jump in that round, before A, C, and E take their third jump.\n\nWhen the total number of jumpers is 9 or less, the competition reverts to a single rotation of all 9 jumpers.