Question

Prove this identity: \(\mathrm{P}(\mathrm{n}, \mathrm{n}-1)=\mathrm{P}(\mathrm{n}, \mathrm{n})\).

Short answer

We can prove the identity \(P(n, n-1) = P(n, n)\) using the permutations formula \(P(n, k) = \frac{n!}{(n-k)!}\). By applying the formula, we get: \(P(n, n-1) = \frac{n!}{1!} = \frac{n!}{1}\) and \(P(n, n) = \frac{n!}{0!} = \frac{n!}{1}\). Since both expressions are equal, the identity is proved: \(P(n, n-1) = P(n, n)\).

Step-by-step solution

Write down the given identity

The given identity to be proved is: \[P(n, n-1) = P(n, n)\]

Write the formula for permutations

The formula for permutations is: \[P(n, k) = \frac{n!}{(n-k)!}\]

Apply the formula to both sides of the given identity

We will now apply the formula for permutations to both sides of the given identity: \[P(n, n-1) = \frac{n!}{(n-(n-1))!}\] and \[P(n, n) = \frac{n!}{(n-n)!}\]

Simplify both sides

Let's simplify the expressions on both sides of the equation: For the left side: \[P(n, n-1) = \frac{n!}{(n-(n-1))!} = \frac{n!}{1!}\] For the right side: \[P(n, n) = \frac{n!}{(n-n)!} = \frac{n!}{0!}\] Recall that 0! is defined as 1. Therefore, we get: \[P(n, n) = \frac{n!}{1}\]

Compare both expressions and conclude

After simplification, we get the following expressions: \[P(n, n - 1) = \frac{n!}{1}\] and \[P(n, n) = \frac{n!}{1}\] Since both expressions are equal, we can conclude that the given identity is true: \[P(n, n-1) = P(n, n)\]. Hence, the identity has been proved.

Key concepts

Factorial Notation

Factorial notation plays a crucial role in permutations and combinatorics. It's denoted by an exclamation mark (!) after a number and means the product of all positive integers from that number down to 1. For example, the factorial of 5, written as \(5!\), is calculated as \(5 \times 4 \times 3 \times 2 \times 1\), which equals 120.

For zero, the factorial is uniquely defined as \(0! = 1\). This definition might seem odd at first, but it's essential for consistency in various mathematical formulas, including the permutations formula. Understanding factorial notation is key to simplifying and solving permutation problems effectively.

Permutations Formula

The permutations formula allows us to calculate the number of ways to arrange \(k\) objects from a set of \(n\) distinct objects. The formula is stated as \(P(n, k) = \frac{n!}{(n-k)!}\).

This equation signifies that to get the number of permutations, you first obtain the factorial of the total number of objects \(n!\), and then divide by the factorial of the difference between the total and the number of positions to be filled, \((n-k)!\). It's vital to note that order is important in permutations—changing the order of the items results in a different permutation.

Mathematical Proofs

Mathematical proofs are logical arguments that demonstrate the truth or falsity of a given statement. They follow a sequence of steps that must be both valid and sufficient to establish the statement's conclusion.

When conducting a proof, such as proving an identity, it's essential to start with known formulas or axioms. Each subsequent step builds on the previous ones, ensuring that the reasoning process is sound. In our example, the proof moves systematically from stating the identity to breaking it down into known permutations formulas before simplifying the expressions to illustrate equivalence.

Simplifying Expressions

Simplifying expressions involves reducing them to their most basic form without changing their value. In mathematics, particularly in algebra and arithmetic, simplification can involve factoring numbers, canceling common terms, and applying known identities like \(0! = 1\).

When simplifying expressions in permutations problems, it's crucial to recognize opportunities to simplify factorial expressions. For example, in \( \frac{n!}{(n-(n-1))!} \) by understanding that \(n - (n - 1) = 1\), we can quickly simplify to \(n!\) since \(1! = 1\). These simplifications are key steps in proofs and help to draw clear conclusions.