The Knockout Problem Solution is shown below
366 players participate in a knock-out tournament. In each round, all competing players pair together and play a match, the winner of each match moving to the next round. If at the end of a round, there is an odd number of winners, the unpaired one moves to the next round without playing a match.What is the total number of matches played ?
The solution is given below
So, initially we have 366 players. Since this is an even number, we can have 183 pairs and hence 183 matches.
Now we have 183 winners from the held matches. However now 183 is an odd number.
So 91 pairs can be formed, but one person will be left alone. Hence 91 matches will be played and the lone person will advance to the next round.
Now we have 91 winners + 1 person who advanced = 92 players.
Now this being an even number again, we can have 46 pairs of players, hence 46 matches being played
Now we get 46 winners. Once again, 23 pairs are made and 23 matches are played.
Now since 23 is an odd number, like before 11 pairs of players will be made and the remaining guy will advance.
So we’ll have 11 winners from 11 matches +1 person who advanced = 12.
This will again break down to 6 matches giving us 6 winners.
These will be made into 3 pairs giving us 3 more matches.
Now the final 3 remain. The first 2 will form a pair and play 1 match. Player 3 will automatically advance.
Then player 3 will play with the winner of this match, thus giving us the 1 final match!
Hence if we count all the matches played, we get :
183 + 91 + 46 +23 + 11 + 6 + 3 + 1 +1 = 365
Hence the answer is (c) 365
If you have any doubts, leave me a comment below and Ill get back to you!