Question

In chess championship 153 games have been played. If a player with every other player plays only once, then the number of players are (a) 17 (b) 51 (c) 18 (d) 35

Short answer

The number of players in the chess championship is \(18\). Therefore, the correct answer is (c) 18.

Step-by-step solution

Understand the concept of combinations

In combinatorics, a combination is a selection of items from a larger set, such that the order of the items does not matter. In this case, we are looking for the number of unique pairs of players that can be formed from the total number of players. The formula for combinations is: \[C(n, k) = \frac{n!}{k!(n-k)!}\] Where \(n\) is the total number of items (in our case, the number of players), \(k\) is the number of items in each combination (in this case, 2, since we are looking for unique pairs), and \(C(n, k)\) is the number of combinations.

Find the number of players corresponding to the number of games played

Given that 153 games have been played, our goal is to find a value of \(n (the number of players)\) such that \(C(n, 2) = 153\), using the combinations formula from step 1. Rewriting the equation with the formula, we get: \[\frac{n!}{2!(n-2)!} = 153\] We need to find the value of \(n\) that satisfies this equation.

Solve for n

We will now solve the equation for \(n\): \[\frac{n!}{2!(n-2)!} = 153\] We know that \((n-1)! = \frac{n!}{n}\), so we can rewrite the equation as: \[\frac{n(n-1)}{2} = 153\] Now, multiply both sides by 2: \[n(n-1) = 306\] Since 17 * 18 = 306, \(n = 18\).

Select the correct answer

The correct answer is the number of players who participated in the chess championship: 18 players. Thus, the answer is (c) 18.

Key concepts

Understanding Combinations

Combinations are a fundamental concept in combinatorics. They refer to the selection of a set of items from a larger group, where the order does not matter. This is different from permutations, where order is significant. In our chess championship problem, combinations are used to determine how many unique pairs of players can be made from all participating players. In mathematical terms, this is given by the formula:

• \[C(n, k) = \frac{n!}{k!(n-k)!}\]

Here, \(n\) represents the total number of players, and \(k\) is the size of each group we want to form, which is a pair in this context. We are looking for \(C(n, 2)\) because each game involves a pair of players. So, combinations help to find potential pairs when order does not matter, making them perfect for counting non-repetitive interactions like games.

Deciphering Permutations

While combinations focus on selection without regard to order, permutations involve arrangements where the order does matter. This concept is useful when we need to know in how many different ways we can arrange a set, rather than just selecting items. It is expressed with the formula:

• \(P(n, k) = \frac{n!}{(n-k)!}\)

In the context of our chess championship, permutations would be concerned with sequences or arrangements of playing games, which are not typically how we count game pairings in this scenario since only pair formation is considered. However, recognizing when to use permutations is useful in problems where order or arrangement of participants or items is crucial.

What Are Factorials?

Factorials are a crucial part of both combinations and permutations calculations. The factorial of a number, denoted by \(n!\), is the product of all positive integers up to \(n\). For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials help in expanding and simplifying the formulas for combinations and permutations:

• In combinations \(C(n, k) = \frac{n!}{k!(n-k)!}\), factorials in the denominator cancel out large parts of the numerator, simplifying calculations.
• In permutations \(P(n, k) = \frac{n!}{(n-k)!}\), factorials account for all possible arrangements within a selection.

Utilizing factorials makes complex calculations manageable, allowing us to compute possible groupings and arrangements effortlessly. Understanding how they break down large numbers into a product of sequences provides insights into solving many combinatorial problems.