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if there are 12 teams in a basketball tournament and each team must play every other team in the eliminations how many elimination games will there be?

Sagot :

rylai04

Answer:

66

Step-by-step explanation:

Basketball is a game played between 2 teams. The problem is asking how many total games will there be. This can be seen as taking 2 teams to play from 12 possible teams. We don't care about the order, but we care about one team playing another team.

A combination is a selection of items from a collection such that the order of the selection does not matter. The formula for taking r things from n possible ones with no particular order is expressed by:

[tex]\frac{n!}{n!(n-r)!}[/tex]

The "!" symbol is called "factorial". In simple terms,

[tex]x! = (x)(x-1)(x-2)...(3)(2)(1).[/tex]

For example, 4! is 4*3*2 = 24. Since it is a treated product, we can cancel out fractions with both x! on the numerator and denominator.

There are 2 ways to solve this problem.

This first way is via combination. We have n = 12, r = 2.

Substituting n = 12, r = 2 to the combination formula gives us:

[tex]\frac{n!}{n!(n-r)!}\\\\\frac{12!}{2!(12-2)!}\\\\\frac{12!}{2!10!}\\\\\frac{12*11*10!}{2*10!}\\\\6*11\\\\66[/tex]

There are 66 possible games in the 12-team tournament.

The second way is a bit intuitive. The first team, team 1, will play 11 of the teams, from team 2 to team 12. The second team, having played team 1 already, only plays 10 of the teams, from team 3 to team 12. This goes on until Team 11 plays 1 team, Team 12.

We can add the following games.

[tex]11+10+9+8+7+6+5+4+3+2+1 = 66[/tex]

We got the same answer from earlier, 66.

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