Problem 11
Question
Evaluate each factorial expression. \(\frac{9 !}{6 !}\)
Step-by-Step Solution
Verified Answer
504
1Step 1: Understand the factorial function
A factorial of a number n, denoted as n!, is the product of all positive integers less than or equal to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1\).
2Step 2: Write the given expression with factorials expanded
The given expression is \( \frac{9!}{6!} \). Expand the factorials to see the pattern: \[ \frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \]
3Step 3: Simplify the common terms
Observe that \(6!\) is common in both the numerator and denominator. Cancel out these common terms: \[ \frac{9 \times 8 \times 7 \times \boxed{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{\boxed{6 \times 5 \times 4 \times 3 \times 2 \times 1}} = 9 \times 8 \times 7 \]
4Step 4: Calculate the remaining product
Multiply the remaining numbers: \[ 9 \times 8 = 72 \] and then, \[ 72 \times 7 = 504 \]
Key Concepts
Factorial FunctionSimplifying ExpressionsNumerical Multiplication
Factorial Function
The factorial function is an essential tool in mathematics. It helps in various areas like algebra, probability, and combinatorics. To understand a factorial function, consider the notation given as n!. This symbol stands for the product of all positive integers up to n.
For example, if n is 5, then the factorial of 5, written as 5!, is calculated as:
\( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factoring is also used in simplifying expressions involving fractions and large numbers. Always remember that:
For example, if n is 5, then the factorial of 5, written as 5!, is calculated as:
\( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factoring is also used in simplifying expressions involving fractions and large numbers. Always remember that:
- Factorials grow very fast.
- They are only defined for non-negative integers.
- \(0!\) is defined as 1.
Simplifying Expressions
Simplifying expressions involving factorials makes calculations much easier. Let's take the example given in the exercise:
We need to simplify the expression \(\frac{9!}{6!}\).
First, expand both the numerator and the denominator:
\[ \frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \].
Notice that \(6!\) is present in both the numerator and the denominator. This means they will cancel each other out.
After canceling the common terms, you're left with:
\[ 9 \times 8 \times 7 \].
This simplification greatly reduces the complexity of the problem and makes numerical calculations easier.
We need to simplify the expression \(\frac{9!}{6!}\).
First, expand both the numerator and the denominator:
\[ \frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \].
Notice that \(6!\) is present in both the numerator and the denominator. This means they will cancel each other out.
After canceling the common terms, you're left with:
\[ 9 \times 8 \times 7 \].
This simplification greatly reduces the complexity of the problem and makes numerical calculations easier.
Numerical Multiplication
Numerical multiplication is another fundamental skill for evaluating factorial expressions. In our example, once we've simplified the expression to \(9 \times 8 \times 7\), we proceed to multiply these numbers.
First, multiply 9 by 8:
\[ 9 \times 8 = 72 \].
Next, take the result and multiply by 7:
\[ 72 \times 7 = 504 \].
This gives us the final answer:
\( \frac{9!}{6!} = 504 \).
Practicing multiplication with smaller numbers builds confidence and accuracy before moving on to larger numbers and more complex operations.
First, multiply 9 by 8:
\[ 9 \times 8 = 72 \].
Next, take the result and multiply by 7:
\[ 72 \times 7 = 504 \].
This gives us the final answer:
\( \frac{9!}{6!} = 504 \).
Practicing multiplication with smaller numbers builds confidence and accuracy before moving on to larger numbers and more complex operations.
Other exercises in this chapter
Problem 11
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