For someone who doesnt know about combination formulae:
consider 4 teams A B C D
Listing down number of matches for these 4 teams as per the given conditions:
AvA - a team cannot play against itself
AvB
AvC
AvD
Number of matches:
3BvA - Already counted above
BvB - a team cannot play against itself
BvC
BvD
Number of matches :
2CvA - Already counted above
CvB - Already counted above
CvC - a team cannot play against itself
CvD
Number of matches :
1DvA - Already counted above
DvB - Already counted above
DvC - Already counted above
DvD - a team cannot play against itself
Number of matches :
0Total number of matches: 3+2+1+0=
6For 4 teams, number of matches as per given conditions is: 3+2+1 i.e (4-1)+(4-2)+(4-3)+(4-4)=3+2+1+0=6
You can calculate for any number of teams using this approach: Use
Arithmetic Progression of
common difference 1 for a higher number of teams as
First term will be (Number of teams-1) and
last term would be 1For the given problem:Total number of matches= 11+10+9+...+1 = {(11+1)*11}/2 = 66