At. The first meeting of a chess club the 9 members decide to have a tournament in which each player plays a game against every other player. How many games will there be??

Can you help me figure this out ?

Each of the 9 players would play against 8 others, that would give us 9x8 or 72

BUT, wouldn't that include such games as
A vs B and B vs A, which is the same game.
So we have to divide by 2 to get all those duplications taken out
number of games = 72/2 = 36

In mathematical symbols that would be
C(9,2) = 9!/(2!7!) = 36

John has five more than twice the number of CDs that Kate has. Together they have 100 CDs

Of course! To determine the number of games in a round-robin tournament where each player plays against every other player, you can use a combination formula.

In this case, there are 9 members in the chess club. To find the number of games, we need to find the number of all possible pairs of these 9 members.

The formula for combinations is:

C(n, r) = n! / (r! * (n-r)!)

In our scenario, n represents the total number of players (9), and r represents the number of players in each game (2, since it's a one-on-one game). The exclamation mark (!) denotes the factorial of a number.

Now let's calculate the number of games:

C(9, 2) = 9! / (2! * (9-2)!)
= 9! / (2! * 7!)

Using factorials, we simplify further:

= (9 * 8 * 7!) / (2 * 1 * 7!)
= (9 * 8) / 2
= 72 / 2
= 36

Therefore, there will be a total of 36 games in the tournament.

Let me know if there's anything else I can help you with!