Misc Funda

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  Tournament Funda There are 16 teams and they are divided into 2 pools of 8 each. Each team in a group plays against one another on a round-robin basis. Draws in the competition are not allowed. The top four teams from each group will qualify for the next round i.e round 2. In case of teams having the same number of wins, the team with better run-rate would be ranked ahead. 1. Minimum number of wins required to qualify for the next round _____? 2. Minimum number of wins required to guarantee qualification in the next round _____? Now, i don't know how many of you are aware of the following method. But 1 thing I mention in advance that this should take only 30 seconds to solve 1. 1 group is consisting of 8 teams. So each team will play 7 match each. Suppose each of the 8 teams were seeded and we consider the case where a higher seeded team will always win. So the number of wins for the 8 teams would be 7,6,5,4,3,2,1,0 with highest seeded team winning all and lowest seeded team losing all. For minimum number of wins we allow 3 teams to win maximum number of matches. Of the remaining 5 teams just find out the mean of their number of wins. In this case it would be (4+3+2+1+0)/5=2. So 5 teams can end up with 2 wins each and a team with better run rate will qualify with 2 wins. 2. In this case consider the mean of first 5 higher seeded teams (7+6+5+4+3)/5=5 So it may be the case that 5 teams can end up having 5 wins each. And hence 1 team will miss the second round birth. So minimum number of wins to guarantee a place would be 6. So guys whenever a similar questions appears you should not be taking more than 30 seconds.  Round Robin/League Games: Suppose A,B,C,D,E,F play each other exactly once. Top 2 Qualify for Finals. 4 Standard Questions: Q1. Find total no. of matches. Q2. Min. no. of matches a team must win in order to have a chance of qualifying. Q3. Max. no. of matches a team can win and still not qualify. Q4. Min. no. of matches a team must win to reach next round undisputed/ guaranteed. A1. Easiest of the lot. A2. If top k teams are supposed to qualify for the next round then choose top k-1 teams and make them win as many matches as possible. Distribute the remaining points equally/ as equal as possible. A3. If top k teams are supposed to qualify then choose top k+1 teams and make them beat every other team then you'll be left with matches between k+1 teams. Distribute as equally as possible. A4. A3 + 1 For the example quoted. A1. 6C2 A2. A 5 B 2 C 2 D 2 E 2 F 2 Answer is 2. A3. A 4 B 4 C 4 Answer is 4 A4. Answer is 5.  UNDERSTANDING PnC, difference between identical and distinct 1. how many ways one can put (a) 4 different boll in 4 different bag (b) 4 identical ball in 4 different bag (C)4 different boll in 4 identical bag (d)4 identical boll in 4 identical bag 1) 4 identical balls in 4 identical bags: (This is easy one) 5 ways: (4, 0, 0, 0), (3, 1, 0, 0), (2, 2, 0, 0), (2, 1, 1, 0) or (1, 1, 1, 1) (2) 4 identical balls in 4 different bags: (This is also easy one) It is like this: a+b+c+d = 4 where a, b, c and d are number of balls in each bag. So, 7C3 ways (3) 4 different balls in 4 different bags: (This is also easy one) Each ball can go to one of the 4 bags. So, it has 4 choices. So, it will be 4*4*4*4.. 4 choices for each. So, 4^4. (4) 4 different balls in 4 same bags: (This is calculation bsed..) This is based on first case (1) (4, 0, 0, 0) => 1 way only (3, 1, 0, 0) => 4C3 ways  3 balls to be grouped can be chosen in 4C3 ways and remaining one in 1 way (2, 2, 0, 0) => 2 balls can be chosen in 4C2 ways. So, 6 ways. But, as we select 2, at the same time we make one more pair of remaing 2. So, at one time, we take care of 2 cases. So, we can take 4C2/2 = 3 ways. (2, 1, 1, 0) => 4C2 4C2 ways to chose 2 balls and then reaming will be 1 and 1. (1, 1, 1, 1) => 1 way only
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