Problem 14 - Discrete Mathematics with Applications

Question

Mark each sentence as true or false, where \(n\) is an arbitrary nonnegative integer and \(0 \leq r \leq n\). $$P(n, 1)=P(n, n-1)$$

Step-by-Step Solution

Answer

The given statement \( P(n, 1)=P(n, n-1) \) is false, as \( P(n, 1) = n \) and \( P(n, n-1) = n! \), and in general, \( n \neq n! \).

Step 1: Understand the Permutations Formula

Permutations refer to the different ways one can arrange items from a given set. The formula for calculating permutations with repetition is: \( P(n, r) = \frac{n!}{(n-r)!} \), where n is the total number of items, r is the number of items being chosen to arrange, and ! denotes the factorial function. Factorial is the product of all positive integers less than or equal to the number (e.g., n! = n × (n-1) × (n-2) × ... × 1).

Step 2: Calculate P(n, 1)

According to the permutations formula, we can calculate \(P(n, 1)\) as follows: \( P(n, 1) = \frac{n!}{(n-1)!} \) Since \( (n-1)! = n! / n \), we get: \( P(n, 1) = n \)

Step 3: Calculate P(n, n-1)

We can now calculate \(P(n, n-1)\) using the same formula: \( P(n, n-1) = \frac{n!}{(n-(n-1))!} \) \( P(n, n-1) = \frac{n!}{1!} \) Given that 1! = 1, we can simplify the expression: \( P(n, n-1) = n! \)

Step 4: Compare P(n, 1) and P(n, n-1)

Now let's compare the results from Step 2 and Step 3: \( P(n, 1) = n \) \( P(n, n-1) = n! \) Since \( n \neq n! \) in general, we can conclude that the given statement is false. #Conclusion# The given statement is false: \( P(n, 1) \neq P(n, n-1) \)

Key Concepts

Factorials

In mathematics, a factorial is an operation that expresses the product of an integer and all positive integers below it. It is denoted with an exclamation mark (!). Factorials are essential in permutations, combinations, and many other mathematical concepts. For example, the factorial of 5, expressed as 5!, would be calculated as:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Factorials grow very fast in value. Even though they start quite modestly (like 1! = 1 and 2! = 2), by the time you reach 10!, the value explodes to 3,628,800.
Factorials are a fundamental building block for calculating permutations. For instance, when arranging the entire set of n distinct items, you would compute n! to find how many different specific sequences are possible.

Permutations Formula

The permutations formula is a key concept in combinatorics for counting the number of possible arrangements of a subset of items from a larger set. The formula for permutations without repetition is:

• \( P(n, r) = \frac{n!}{(n-r)!} \)

Here, \( n \) is the total number of items, and \( r \) represents how many items you want to arrange. This formula calculates how many different sequences can result when selecting \( r \) items from \( n \).
For example, in the problem addressed, when calculating \( P(n, 1) \), the permutation is computed for arranging one item, giving a straightforward result since there are \( n \) options for this single selection.
On the other hand, \( P(n, n-1) \) involves arranging \( n-1 \) items out of \( n \). This requires dividing the factorial of \( n \) by the factorial of the difference, resulting in \( n! \), illustrating a complete set of variations.

Combinatorics

Combinatorics is a branch of mathematics concerned with counting, arranging, and analyzing configurations of sets of objects. It encompasses various concepts including permutations, combinations, and more. This field studies both the existence of certain configurations and the enumeration of distinct arrangements possible amongst items.
Within combinatorics, permutations focus on the ordering of elements, which is what the problem in question emphasizes. This means permutations take into account sequence importance, contrary to combinations which consider selections without regard to order.

• Permutations: focus on the order.
• Combinations: focus on selection, order doesn't matter.

The exercise primarily engages with the permutation side of combinatorics, addressing how different sequences can arise from selecting and organizing subsets of a set of items. Understanding permutations is crucial for solving advanced problems in combinatorics, probability, and specific real-world applications such as scheduling and cryptography.