Question

Problem-Solving Application (15 Points). Eight people enter a racquetball tournament in which each person must play every other person exactly once. Determine the total number of games that will be played? Solve the problem using George Polya's four-step solution Understanding the problem: Devising a plan Carrying out the plan Looking Back

Answer 1

Understanding the problem:

Each of 8 persons play with every other person, i.e., other 7 persons. So, each person plays 7 games.

If one person played with the other person, it implies that the other person also played with that one person and will not play again because each person must play with the other person only once.

A person cannot play with himself/herself.

Devising a plan:

Let 8 persons be A, B, C, D, E, F, G, H

A plays with B, C,.... and H.

B plays with A, C, D,.... and H.

C plays with A,... and H. (except with C, i.e., himself/herself)

D plays with A,..... and H. (except with D)

E plays with A,..... and H. (except with E)

F plays with A,..... and H. (except with F)

G plays with A,..... and H. (except with G)

H plays with A,..... and H. (except with H)

Carrying out the plan:

So, each of 8 persons play with 7 other persons (other than himself/herself).

So, total number of games seem to be 8*7 =56

But when we say, A played with B, it implies B also played with A. So, that becomes only one game, not two. But in the above calculation of 8*7, we included each game twice. So, we need to divide 8*7 by 2.

So, total number of games that will be played =(8*7)/2 =56/2 =28

Looking back:

If we want to know what are the 28 games (i.e., without counting each game twice), here it is:

A plays with B, C,.... and H (7 games)

B plays with C, D,.... and H (6 games) (here, we did not consider A because when we said A played with B, B also played with A and apply the same to the remaining persons below).

C plays with D,... and H (5 games)

D plays with E,... and H (4 games)

E plays with F, G and H (3 games)

F plays with G and H (2 games)

G plays with H (1 game)

So, total number of games that will be played =7+6+5+4+3+2+1 =28

(OR) to solve 1+2+....... +7 we have n =7 natural numbers(1^st 7 natural numbers) and their sum =n(n+1)/2 =7(7+1)/2 =7(8)/2 =28 games.