In a season of football, every team plays every other team exactly one time. If there are 105 games played in one season, how many teams are in the league?

Question

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In a season of football, every team plays every other team exactly one time.

If there are 105 games played in one season, how many teams are in the league?

Answer 1

Let there be N teams in the league.  Then each of N teams plays (N-1) games ["plays every other team exactly once"].  It takes two teams to play a game, so the number of games is:

   N(N-1)/2   (count games played against each other once, not twice).

 

The number of games is given as 105.  Solve for N.
 
N(N-1)/2 = 105
N² - N - 210 = 0
 N = 15         (either factor or use quadratic formula)

 

Note:  This is Gauss' very important formula for the sum of the numbers from 1 to (N-1).

Answer 2

this is the reverse of the problem: if there are, say, 8 people at a party, and everybody shakes hands with everybody else exactly once, how many handshakes will there be ?

there are two ways to solve this; use simple arithmetic or use the combination formula: choose 2 from 8 in how many ways ?

if there are 105 games played, how many teams are there ?

n=number of teams

choose 2 from n teams in how many ways ??

n!/2!(n-2)!=105

n!/2(n-2)!=105

n!=2(n-2)!(105)

n!=(n-2)!(210)

n(n-1)(n-2)(n-3)(n-4)...=210(n-2)(n-3)(n-4)...

cancel the n-2, n-3, n-4, and so on...on both sides

n(n-1)=210

n^2-n=210

n^1-n-210=0

210=2*3*5*7=(2*7)*(3*5)=14*15 and 15-14=1

factor

(n-15)(n+14)=0

n-15=0

n=15 teams

check: 15!/2!(15-2)!=15!/2(13)!=15*14/2=210/2=105

or,...

the first team plays 14 other teams

the second team plays 13 other teams

the third team plays 12 other teams

and so on...

14+13+12+11+10=9+8+7+6+5+4+3+2+1=105