Problem 33 - Algebra and Trigonometry with Analytic Geometry

Question

A committee is going to select 30 students from a pool of 1000 to receive scholarships. How may ways could the students be selected if each scholarship is worth (a) the same amount? (b) a different amount?

Step-by-Step Solution

Answer

(a) \( \binom{1000}{30} \) ways; (b) \( P(1000, 30) \) ways.

Step 1: Understand the Problem

We need to find two separate answers based on different conditions for scholarship selection from the pool of 1000 students. The conditions differ based on whether the scholarships are identical or differentiated by amounts.

Step 2: Calculate for Identical Scholarships

When the scholarships are identical, we only care about which 30 students are selected from the 1000. This is a combination problem. The number of ways to choose 30 students from 1000 is given by the combination formula:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Here, \( n = 1000 \) and \( r = 30 \). Thus, the number of ways is:\[ \binom{1000}{30} = \frac{1000!}{30! \, 970!} \]

Step 3: Calculate for Different Scholarships

When each scholarship has a different amount, the order of selection matters, as each selected student will receive a different scholarship. Thus, this scenario uses permutations. The number of ways to choose and order 30 students from 1000 is given by the permutation formula:\[ P(n, r) = \frac{n!}{(n-r)!} \]Where \( n = 1000 \) and \( r = 30 \). Therefore, the number of ways is:\[ P(1000, 30) = \frac{1000!}{970!} \]

Step 4: Plug in the Numbers

The exact values for these computations aren't typically calculated by hand due to their size, but we use a calculator or software to evaluate them. The combination \( \binom{1000}{30} \) results in a very large number, and the permutation \( P(1000, 30) \) will be even larger, reflecting the additional complexity of considering order.

Key Concepts

Permutations

Permutations are a fundamental concept in combinatorics dealing with the arrangement of objects. When we talk about permutations, we are interested in the order of the selection. This means if you have a different arrangement of the same group of items, it counts as a separate permutation.
For example, if we're considering giving 30 students scholarships that each have different amounts, who gets which scholarship matters. Therefore, we use permutations in such a context.

• Formula: The formula to determine how many permutations are possible when selecting \( r \) items from \( n \) items is: \[P(n, r) = \frac{n!}{(n-r)!}\]
• Explanation: Here, \( n! \) (n factorial) represents the product of all positive integers up to \( n \). It's used because we're ordering all choices, and then divided by \((n-r)!\) to remove excess arrangements.

Using the permutation formula, we calculated the number of ways to arrange the scholarships when each is of different value.

Combinations

Combinations are another key concept in combinatorics, often contrasted with permutations. Unlike permutations, combinations do not consider the order of selection. When we care only about which items are selected and not the sequence in which they're selected, we use combinations.
In the case of scholarships being identical, any selection of 30 students from a pool of 1000 is considered the same, regardless of order. Thus, we use combinations to calculate this.

• Formula: The formula for combinations when selecting \( r \) items from \( n \) items is: \[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]
• Explanation: This formula calculates the total number of groups that can be selected by factoring out the arrangements that are repeated because order does not matter.

Using combinations, we found the possibilities for selecting students when the scholarships are identical.

Scholarships

Scholarships in this context serve a dual purpose: as a means of financial assistance and as a mathematical tool to understand selection problems. The problem can be simplified using combinatorics, where scholarships are treated as items to distribute among students.
When scholarships are of the same value, it translates mathematically to a combination since who receives which scholarship doesn't matter. However, if each scholarship is of a unique value, it becomes a permutation problem, adding an order component to the selection. This highlights the application of permutations and combinations in understanding real-life selection scenarios.
Understanding scholarships through these mathematical concepts helps in grasping how different scenarios can completely alter the approach to solving the problem.

Selection Process

The selection process is crucial in determining how students receive scholarships. Mathematically, it involves choosing a subset from a larger set, often requiring an understanding of whether order is essential or not. This decision affects whether permutations or combinations are used.

• In the exercise, if scholarships are identical, the selection process is straightforward as order doesn't matter, leading to a combination calculation.
• Conversely, if each scholarship had a different value, the process considers the arrangement, hence requiring permutations.

For any selection process, it’s key to distinguish between scenarios where order matters and where it doesn’t, as this influences the mathematical approach used to determine the total possible outcomes. This clarity ensures the correct application of formulas, giving us insight into the complexity of the selection involved.