Twenty students hold a chess tournament in which each of the students will play every other student in the group once. How many individual contest will be held?

Question

Twenty students hold a chess tournament in which each of the students will play every other student in the group once. How many individual contest will be held?

Answer 1

190 contests

This is a combination. A combination is written as either _nC_r or (ⁿ_r). These are similar to permutations, except order does not matter, so you must divide by the number of ways they can be reordered. As a formula, it looks like this: _nC_r=n!/((n-r)!r!)As you can see, the only difference between this and the normal permutation without repetition formula is the r!, which gives the possible rearrangements for the group size. Therefore, this combination can be written as ₂₀C₂ or (²⁰₂)=20!/(18!2!).

 20! and 18! cancel to make (20x19)/(2)=10x19=190. This can also be done on a graphing calculator by pressing the math button, going to the PRB tab and selecting _nC_r. You must enter n first, then select the function on the calculator, then enter r and press enter. Hope this was helpful!

Answer 2

Call the 20 students A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S, and T

A plays the 19 other students

B plays C,D,E,... for a total of 18 games(he doesn't play A because A played B and they play only once)

C plays D,E,F,... for a total of 17 games(he doesn't play A or B because A and B already played C)

D plays E,F,G,... for a total of 16 games(above reasoning)

E plays F,G,H,... for a total of 15 games(above reasoning)

F plays G,H,I,...  for a total of 14 games(above reasoning)

Do you see the pattern ?

G plays 13 games

H plays 12 games

I plays  11 games

J plays  10 games

K plays 9 games

L plays 8 games

M plays 7 games

N plays 6 games

O plays 5 games

P plays 4 games

Q plays 3 games

R plays 2 games

S plays 1 game

T plays no more games because everybody else already played T

19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1=190 contests