91Ó°ÊÓ

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 \(n(n-1)\).

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 \(\frac{2^k(2k-1)!}{(k-1)!k!}\).