Problem 33 - Algebra and Trigonometry

Question

These problems involve permutations. Contest Prizes In how many ways can first, second, and third prizes be awarded in a contest with 1000 contestants?

Step-by-Step Solution

Answer

There are 997,002,000 ways.

Step 1: Understand the Problem

The problem requires finding the number of ways to award three prizes (first, second, and third) among 1000 contestants. Each position must have a different contestant, implying that order matters. This is a permutation problem because the order of the prize winners is important.

Step 2: Define the Permutation Formula

The number of permutations of choosing k objects from a set of n objects is given by the formula \[ P(n, k) = \frac{n!}{(n-k)!} \]where \(n\) is the total number of objects, and \(k\) is the number of objects to be chosen.

Step 3: Apply the Formula

For this problem, we have \(n = 1000\) contestants and \(k = 3\) positions (first, second, and third prizes). Plug these values into the permutation formula:\[ P(1000, 3) = \frac{1000!}{(1000-3)!} = \frac{1000 \times 999 \times 998}{1} \]

Step 4: Calculate the Result

Compute the above expression:\[ P(1000, 3) = 1000 \times 999 \times 998 = 997002000 \]

Key Concepts

Permutation Formula

In combinatorics, a permutation relates to the different ways an arrangement of objects can be structured, particularly when the order of selection matters. The permutation formula is a handy tool to calculate such arrangements. It's given by:\[ P(n, k) = \frac{n!}{(n-k)!} \]Let's break down this formula:

• \(n!\) (n factorial) represents the product of all positive integers up to \(n\). Thus, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
• \((n-k)!\) is the factorial of the difference between the total number \(n\), and the subset you're selecting, \(k\). This helps in calculating permutations when not all objects are used.

So, whether arranging books on a shelf or awarding prizes, this formula is your go-to for situations where every different order counts.

Combinatorial Problems

Combinatorial problems often require us to count arrangements or groups of objects under certain conditions. They can include permutations or combinations, depending on whether the order of selection matters. Permutations are used when:

• Each selected option is unique and cannot be repeated.
• The sequence in which the items are arranged is crucial.

In the contest problem, the order of winners for the first, second, and third prizes is essential, making it a permutation problem. Unlike combinations, which focuse on choosing groups regardless of order, permutations give us the total number of possible arrangements respecting positional differences. This makes them ideal for problems where rank or sequence is crucial.

Order of Selection Importance

Understanding when the order of selection is important can drastically affect the way in which a problem is solved, particularly in combinatorial settings. In essence, the order matters when:

• The position each selected object holds is crucial to the outcome.
• Changing the order results in a different arrangement or scenario.

In our contest example, awarding the first, second, and third prizes necessitates differentiating between the positions – swapping names changes the outcome. This contrasts with situations where the grouping is all that counts, such as assembling a team without specified roles. Here, permutations rule because each sequence creates a unique set of outcomes, each significant in its own right. Recognizing this helps in distinguishing when to apply permutations over other combinatorial principles.