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 ?

a). 366

b). 282

c). 365

d). 418

The solution is given below

**EXPLANATION :**

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!