Question

For every positive integer n greater than 1, n! is defined as the product of the first n positive integers. For example, \(4! = (1)(2)(3)(4) = 24\). What is the value of \(\frac{12!}{10!}\)?

• A. 2
• B. 66
• C. 121
• D. 132
• E. 144

Hint:

Never try to calculate large factorials fully before simplifying. Always look for cancellations. The expression \(\frac{n!}{k!}\) (for \(n>k\)) simplifies to the product of integers from \(k+1\) up to \(n\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem involves factorials, denoted by the exclamation mark (!). We need to simplify a fraction of two factorials.
Step 2: Key Formula or Approach:
The key to simplifying factorial fractions is to expand the larger factorial until you reach the smaller factorial, which will then cancel out.
We can write \(n! = n \times (n-1) \times (n-2) \times \dots \times 1\), which is also equal to \(n \times (n-1)!\).
Step 3: Detailed Explanation:
We need to calculate \(\frac{12!}{10!}\).
Let's expand the numerator, 12!, using the definition:
\[ 12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]
The denominator is:
\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]
We can rewrite 12! as:
\[ 12! = 12 \times 11 \times (10 \times 9 \times \dots \times 1) = 12 \times 11 \times 10! \]
Now substitute this into the fraction:
\[ \frac{12!}{10!} = \frac{12 \times 11 \times 10!}{10!} \]
The \(10!\) terms in the numerator and denominator cancel each other out.
\[ = 12 \times 11 \]
\[ 12 \times 11 = 132 \]
Step 4: Final Answer:
The value of \(\frac{12!}{10!}\) is 132.