Question

Simplify the following expressions: a. \(\frac{n !}{(n-2) !}\) b. \(\left.\frac{n !}{\left(\frac{n}{2} !\right)^{2}} \text { (for even } n\right)\)

Short answer

The simplified expressions are: a. \(n(n-1)\) b. \(\frac{2^k(2k-1)!}{(k-1)!k!}\)

Step-by-step solution

Expression a: \(\frac{n !}{(n-2) !}\)

Step 1: Break down the factorial in the numerator We can rewrite \(n!\) as \(n(n−1)(n−2)!\) since the definition of a factorial is the product of all positive integers up to that number. Step 2: Substitute the new expression Now we can substitute the new expression for the numerator and simplify. \[ \frac{n(n-1)(n-2)!}{(n-2)!} \newline \] Step 3: Cancel out the \((n-2)!\) Since \((n-2)!\) appears in both the numerator and denominator, we can cancel it out. This leaves us with the simplified expression: \[ n(n-1) \] So the simplified expression for (a) is \(n(n-1)\).

Expression b: $\left.\frac{n !}{\left(\frac{n}{2} !\right)^{2}} \text { (for even } n\right)$

Step 1: Rewrite the expression Rewrite the expression using the fact that \(n\) is even as \(k = n/2\), where \(k\) is an integer. The expression then becomes: \[ \frac{n !}{\left(k!\right)^{2}} \] Step 2: Break down the factorial in the numerator We can rewrite \(n!\) as \(\left(2k\right)\left(2k-1\right)\left(2k-2\right)\cdots\left(2k-2k+3\right)\left(2k-2k+2\right)\left(2k-2k+1\right)\) since \(n = 2k\). Step 3: Rewrite the expression Rewrite the expression as product of two sets of factorials. \[ \frac{\left(2k\right)\left(2k-1\right)\left(2k-2\right)\cdots\left(2k-2k+3\right)\left(2k-2k+2\right)\left(2k-2k+1\right)}{\left(k!\right)^{2}} \] Step 4: Pair every second term Pair every second term in the numerator to get the product of two sets of factorials: \[ \frac{\left[\left(2k\right)\left(2k-2\right)\cdots\left(2k-2k+2\right)\right]\left[\left(2k-1\right)\left(2k-3\right)\cdots\left(2k-2k+1\right)\right]}{\left(k!\right)^{2}} \] Step 5: Simplify the products Identify the products in the numerator as the factorials: \[ \frac{\left(2^k k!\right)\left(\frac{(2k-1)!}{(k-1)!}\right)}{\left(k!\right)^{2}} \] Step 6: Cancel out the \(k!\) terms Cancel out the common terms among the numerators and denominators: \[ \frac{2^k}{k!} \cdot \frac{(2k-1)!}{(k-1)!} \] So, the simplified expression for (b) is \(\frac{2^k(2k-1)!}{(k-1)!k!}\).

Key concepts

Factorial Notation

Understanding factorial notation is crucial for mastering various mathematical concepts, particularly in combinatorics and probability. A factorial is the product of all positive integers from 1 up to a given number. It's denoted by an exclamation point \textbf{( ! )}. For example, the factorial of 4, written as 4!, is calculated as:
\begin{align*}4! = 4 \times 3 \times 2 \times 1 = 24.\text{In the exercise, } n! \text{ represents the factorial of } n \text{. When we want to simplify expressions involving factorials, such as } \frac{n !}{(n-2) !}\text{, we break down the factorial into its components.}\end{align*}
In general, the factorial of any non-negative integer \(n\) is:
\begin{align*}n! = n \times (n-1) \times (n-2) \times \text{...} \times 3 \times 2 \times 1.\text{Additionally, by convention, } 0! \text{ is defined to be 1.}
This notation enables us to simplify expressions by canceling out common factorial terms in both the numerator and denominator, as seen in the step-by-step solution provided.

Algebraic Simplification

Algebraic simplification refers to the process of reducing expressions to their simplest form. Simplification might involve factoring, canceling terms, and applying arithmetic operations. In factorial problems, algebraic simplification often involves breaking down the factorials into their constituent factors and canceling out like terms.
For instance, with the expression \( \frac{n!}{(n-2)!} \), after expanding \(n!\) to \(n(n-1)(n-2)!\), the \((n-2)!\) terms cancel out because they appear in both the numerator and the denominator, leading to the simplified form of \(n(n-1)\).
The simplification process is vital for both manual computations and for understanding underlying mathematical relationships. It can also make complex problems more approachable and easier to solve. When simplifying, always look for opportunities to reduce the expression by cancelling common factors, which was effectively applied in the exercise solution.

Mathematical Problem-Solving

Mathematical problem-solving is a critical skill that combines understanding mathematical concepts, applying algorithms, and logical reasoning to solve problems. In our exercise involving factorial simplification, problem-solving requires a step-by-step approach:

• Identify patterns or properties that can be used (for instance, recognizing that \(n\) is even allows us to rewrite the expression with \(k = \frac{n}{2}\)).
• Break down complex expressions into simpler parts (like turning \(n!\) into a product of accessible terms).
• Utilize algebraic simplification methods, including canceling out common terms to reduce expressions to their simplest form.
• Analyze the revised expression to ensure that all possible reductions have been made.

Problem-solving exercises like simplifying factorials help develop these skills, which are applicable in various areas of mathematics and essential for academic success. In both exercises provided, a methodical simplification process leads to a clear, concise solution, embodying the essence of effective mathematical problem-solving.