The 16 players in a chess tournament must each play every other player

Question

The 16 players in a chess tournament must each play every other player in the first round of the tournament. Only half of the first 16 players will play in the second round, each player playing every other player, and only half of them will play in the third round, a pattern that will continue until one winner is left. How many chess games will be played in the tournament?

(A) 16
(B) 30
(C) 72
(D) 120
(E) 155

(A) 16
(B) 30
(C) 72
(D) 120
(E) 155

BrentGMATPrepNow · Answer

Let's start by calculating the total number of games played by the original 16 players.
In other words, in how many ways can we select two players two play a game of chess?
Since the order in which we select the two players does not matter, we can use combinations. 
We can select 2 players from 16 players in 16C2 ways. 
16C2 = (16)(15)/(2)(1) =  120

Once the first round of play eliminates half the players, we're left with 8 players. 
  Important: Since there are still more games to be played, we know the correct answer will be greater than 120....  
....Answer: E

Kimberly77 · Answer

Hi  BrentGMATPrepNow , I calculate it by listings as below and not sure why is not right?
Player 16 against Player 15 ... 1st game
Player 15 against Player 14 ... 2nd game
Player 14 against Player 13 ... 3rd game
....
Player 2 against Player 1 ... 15th game

So there are 15 chess games will be played in the tournament.

BrentGMATPrepNow · Answer

The question tells us that  The 16 players in a chess tournament   must each play every other player   in the first round of the tournament 
And your solution, Player 16 plays only one game (against Player 15)
In actuality (to follow the conditions set out in the question), Player 16 must play Player 15, Player 14, Player 13, Player 12, . . . . Player 3, Player 2 and Player 1. 
Similarly, Player 15 must play Player 16, Player 14, Player 13, Player 12, . . . . Player 3, Player 2 and Player 1. 
Similarly, Player 14 must play Player 16, Player 15, Player 13, Player 12, . . . . Player 3, Player 2 and Player 1.
As you can see, I've already listed more than 15 games.

Kimberly77 · Answer

Thanks BrentGMATPrepNow , Get it and see why now.

Waqas.Baig · Answer

Hi All, kindly evaluate my approach: based on Brent concept, 4stages will be there as with each stage player is getting halved. So in stage 16player will play against 15others total no of games will be 16*15/2 (divided by 2 as duplicacy is there player 1 playing player 2 and player 2playing against player 1 counted twice). similarly, stage2: 8*7/2 stage 3: 4*3/2 and stage 4: 2*1/2 =155

luisdicampo · Answer

Deconstructing the Question  
There are 16 players.
In each round, every player plays every other player (round-robin).
After each round, only half the players advance, continuing until 1 winner remains.
Key idea: a round-robin with  n  players has  \binom{n}{2}=\frac{n(n-1)}{2}  games.

Step-by-step  
Round 1:  n=16 
 \binom{16}{2}=\frac{16\cdot15}{2}=120

Round 2:  n=8 
 \binom{8}{2}=\frac{8\cdot7}{2}=28

Round 3:  n=4 
 \binom{4}{2}=\frac{4\cdot3}{2}=6

Round 4:  n=2 
 \binom{2}{2}=1

Total games:
 120+28+6+1=155

Answer: E

The 16 players in a chess tournament must each play every other player

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The 16 players in a chess tournament must each play every other player

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