Question

What is the value of each expression? a) \(9 !\) b) \(\frac{9 !}{5 ! 4 !}\) c) \((5 !)(3 !)\) d) \(6(4 !)\) e) \(\frac{102 !}{100 ! 2 !}\) f) \(7 !-5 !\)

Short answer

a) 362880, b) 126, c) 720, d) 144, e) 5151, f) 4920

Step-by-step solution

Calculate the Factorial for a) 9!

The factorial of 9 is calculated by multiplying all positive integers up to 9. \[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \]

Simplify the expression for b) \(\frac{9 !}{5 ! 4 !}\)

Calculate the factorial values for 9, 5, and 4. Then, substitute them into the expression. \[ 9! = 362880, \, 5! = 120, \, 4! = 24 \] Substitute these into the equation: \[ \frac{9!}{5!4!} = \frac{362880}{120 \times 24} = \frac{362880}{2880} = 126 \]

Calculate the expression for c) \((5 !) (3 !)\)

Calculate the factorial values for 5 and 3. \[ 5! = 120, \, 3! = 6 \] Multiply these values together: \[ 5! \times 3! = 120 \times 6 = 720 \]

Calculate the expression for d) \(6 (4 !)\)

First, calculate the factorial of 4. \[ 4! = 24 \] Then, multiply this value by 6: \[ 6 \times 4! = 6 \times 24 = 144 \]

Simplify the expression for e) \(\frac{102 !}{100 ! 2 !}\)

Notice that \(102!\) can be written as \(102 \times 101 \times 100!\). Then use this in the expression: \[ \frac{102 \times 101 \times 100!}{100! \times 2!} \] Cancel \(100!\) and simplify: \[ \frac{102 \times 101}{2!} = \frac{102 \times 101}{2} = 5151 \]

Calculate the expression for f) \(7! - 5!\)

Calculate the factorial values for 7 and 5. \[ 7! = 5040, \, 5! = 120 \] Subtract these values: \[ 7! - 5! = 5040 - 120 = 4920 \]

Key concepts

Factorials

Factorials are a mathematical operation represented by an exclamation mark (!). The factorial of a number is the product of all positive integers up to that number. For example, the factorial of 5, written as 5!, is calculated as:

• 5! = 5 × 4 × 3 × 2 × 1 = 120

Factorials are particularly useful in permutations and combinations, but they also frequently appear in various mathematical expressions and equations.
To simplify the calculations, remember these key factorial values:

• 0! = 1 (by definition)
• 1! = 1
• 2! = 2
• 3! = 6
• 4! = 24
• 5! = 120
• 6! = 720
• 7! = 5040
• 8! = 40320
• 9! = 362880

Understanding how to compute and use factorials will greatly help in simplifying and solving more complex expressions.

Simplifying Expressions

Simplifying expressions involves breaking down complex mathematical equations into simpler, more manageable parts. This process often uses factoring and cancelling terms.
Let's simplify the expression \(\frac{9!}{5!4!}\):

• We first compute the factorial values: \(\begin{align*} 9! &= 362880, \ 5! &= 120, \ 4! &= 24 \end{align*}\)
• Substitute these values into the expression: \(\frac{9!}{5! \times 4!} = \frac{362880}{120 \times 24} = 126\)

This method can also be applied to more complex problems. For example, \(\frac{102!}{100! 2!}\) simplifies by recognizing that \(\frac{102!}{100!} = 102 \times 101!!!! \). The equation becomes \(\frac{102 × 101 × 100!}{100! \times 2!} \), and cancelling \!100! results in \(\frac{102 × 101}{2} = 5151\).
Knowing these steps helps to break down and tackle seemingly difficult mathematical problems.

Arithmetic Operations

Arithmetic operations such as addition, subtraction, multiplication, and division are fundamental for simplifying and calculating expressions.
For instance, consider \((5! \times 3!)\). Calculating the factorials first gives \(\begin{align*} 5! &= 120, 3! &= 6 \end{align*}\). Then, we multiply the results: \(\begin{align*} 5! \times 3! &= 120 \times 6 = 720 \end{align*}\). This sequence of steps ensures accurate results.
Similarly, division problems like \(\frac{9!}{5! 4!}\) require you to divide large numbers efficiently. Knowing shortcut methods or properties, like factorial values, makes this easier.
Finally, in subtraction, such as with \((7! - 5!)\), compute the individual factorials first:

• 7! = 5040
• 5! = 120

Subtract the smaller value from the larger one: 5040 - 120 = 4920. Practicing these operations helps you become more comfortable solving various mathematical problems.