SOLUTION: Can somebody please confirm if my answer is correct? My wild guess is 11. But if I am to use C= 12!/[1!(12-1)!] = 12 is what I get. I will appreciate your response.Thanks. A che

Algebra ->  Probability-and-statistics -> SOLUTION: Can somebody please confirm if my answer is correct? My wild guess is 11. But if I am to use C= 12!/[1!(12-1)!] = 12 is what I get. I will appreciate your response.Thanks. A che      Log On


   

Question 48197: Can somebody please confirm if my answer is correct? My wild guess is 11. But if I am to use C= 12!/[1!(12-1)!] = 12 is what I get. I will appreciate your response.Thanks.
A chess tournament has 12 participants. How many games must be scheduled if every player must play every other player exactly once.
Answer by AnlytcPhil(1810) About Me  (Show Source):
You can put this solution on YOUR website!
Can somebody please confirm if my answer is correct? My wild guess is 11. But
if I am to use C= 12!/[1!(12-1)!] = 12 is what I get. I will appreciate your
response.Thanks. 

A chess tournament has 12 participants. How many games must be scheduled if
every player must play every other player exactly once.

Your wild guess of 11 would be the number of games any one player would be
involved in.  However you are asked how many games are played altogether by
all participants.

The correct answer is

THE NUMBER OF COMBINATIONS OF 12 THINGS TAKEN 2 AT A TIME,
not 1 at a time.

This can either be written as C(12,2) or 12C2

but the formula is

12!/[2!(12-2)!] = 66

Edwin