Question

In a series of races, 10 toy cars are raced, 2 cars at a time. If each car must race each of the other cars exactly twice, how many races must be held?

• A. 40
• B. 90
• C. 100
• D. 180
• E. 200

Hint:

This type of problem is a classic "round-robin tournament" setup. The number of games in a single round-robin tournament with n participants is \(\binom{n}{2}\). If there are multiple rounds, simply multiply by the number of rounds.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a combinations problem. First, we need to determine how many unique pairs of cars can be formed from the 10 cars, as each pair corresponds to a single race. Then, we account for the fact that each race happens twice.
Step 2: Key Formula or Approach:
The number of ways to choose a committee (or pair) of \(r\) items from a set of \(n\) items is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Step 3: Detailed Explanation:
1. Find the number of unique pairings of cars.
We have \(n=10\) cars and we are choosing them in pairs (\(r=2\)). The order in which we choose the two cars for a race does not matter (a race between Car A and Car B is the same as a race between Car B and Car A), so we use combinations.
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} \]
Expand the factorial:
\[ = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} \]
Cancel the \(8!\) term:
\[ = \frac{10 \times 9}{2} = \frac{90}{2} = 45 \]
There are 45 unique pairs of cars, which means 45 unique races are needed for every car to race every other car once.
2. Account for each race being held twice.
The problem states that each car must race each other car "exactly twice".
Total races = (Number of unique races) \(\times\) 2.
Total races = \(45 \times 2 = 90\).
Step 4: Final Answer:
A total of 90 races must be held.