there are 9 fighters and each fighter must fight each of the other fighters once. How many total fights will take place?

Question

no fighters can fight each other more than once but all must fight each other at least once

 

Answer 1

there are 9 fighters

A,B,C,D,E,F,G,H.I

A fights B,C,D,E,F,G,H,and I for a total of 8 fights

B fights C,D,E,F,G,H, and I for a total of 7 fights (B does not fight A because A already fought B)

C fights D,E,F,G,H, and I for a total of 6 fights (C already fought A and B !!)

do you see a pattern ??

D fights E,F,G,H, and I for a total of 5 fights

E fights 4 more people

F fights 3 more people

G fights 2 more people

H fights 1 more person

I already fought everybody else

8+7+6+5+4+3+2+1=36 fights 

you now have an arithmetic sequence with 8 terms and the common difference being -1

S(8)=(8/2)(2*8+[8-1][-1])

S(n)=4(16-7)=4*9=36

if there were 100 fighters, the sequence would be 99, 98, 97, ...,3, 2, 1

to add all these numbers, you would use the formula

also, you can look at the problem as a Combination problem...

how many combinations of 2 fighters are there if you have 9 fighters ?

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

however, looking at the first solution gives you an understanding of the problem and not just the answer