Problem 79 - Precalculus with Limits

Question

In Exercises 77-84, simplify the factorial expression. \( \dfrac{12!}{4! \cdot 8!} \)

Step-by-Step Solution

Answer

\[ \dfrac{3 \cdot 11 \cdot 5 \cdot 9}{7}\]

Step 1: Expanding the Factorials

First, expand out each factorial \\[12! = 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \]\Similarly, \\[4! = 4 \cdot 3 \cdot 2 \cdot 1\] \and \\[8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\]

Step 2: Inserting the Expanded Factorials

Replace each factorial in the given expression with the expanded computations. We have \\[\dfrac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(4 \cdot 3 \cdot 2 \cdot 1) \cdot (8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}\]

Step 3: Cancelling Common Factors

Now we cancel out the common factors that appear in both the numerator and the denominator of our expression. Notice that 4!, which is 4*3*2*1, is part of 8! and also 12!, so we can cancel these factors out directly. This simplification will leave: \\[\dfrac{12 \cdot 11 \cdot 10 \cdot 9}{8 \cdot 7}\]

Step 4: Simplifying Further

Continuing to simplify our expression by cancelling more common terms in the numerator and the denominator, we get: \First, we notice that 12 is divisible by 6 and we can cancel out a 2 in the denominator: \[\dfrac{6 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 7}\]\Then, both 6 and 4 have a common divisible factor of 2, so we cancel, arriving at: \[\dfrac{3 \cdot 11 \cdot 10 \cdot 9}{2 \cdot 7}\]\Finally, both 10 and 2 have a common factor of 2, we cancel that out, and we finally get to: \[\dfrac{3 \cdot 11 \cdot 5 \cdot 9}{7}\]

Key Concepts

Combinatorics

Combinatorics is a fascinating branch of mathematics. It primarily deals with counting, arranging, and combining items effectively. It's like solving a mathematical puzzle to find how many different ways things can be combined or arranged.
For example, combinatorics helps determine the number of possible outcomes when shuffling playing cards or choosing a committee from a group of people.

• Combination: When the order doesn't matter, like selecting team members from a pool.
• Permutation: When the order does matter, like race rankings or seating arrangements.

In this exercise, combinatorics gives us the tools to simplify the expression \(\dfrac{12!}{4!\cdot8!}\). This type of problem is a combinatorial construct often used in binomial coefficients, where you're picking a subset of objects from a larger set.

Permutations

Permutations are all about arrangement, specifically how many ways you can arrange a set of objects. When calculating permutations, the order of the objects is important.
Think of the letters A, B, and C. The permutations of these letters are: ABC, ACB, BAC, BCA, CAB, and CBA. That's 6 different ways, calculated as \(3! = 3 \cdot 2 \cdot 1 = 6\).
In the context of this exercise, simplifying \(\dfrac{12!}{4!\cdot8!}\) involves understanding permutations to reduce the factorial expression efficiently. By dividing factorials such as \(12!\) and \(8!\), you are essentially calculating how many ways we can arrange or pick subsets. Permutations emphasize the significance of each ordering within a set.

Mathematical Notation

Mathematical notation serves as the language of mathematics. It helps us convey concepts, operations, and formulas succinctly.
When dealing with factorials, symbols like \(n!\) tell us to multiply all positive integers up to \(n\). For instance, \(4! = 4 \times 3 \times 2 \times 1\).
In the given problem, \(12!\) stands for the product of all integers from 1 to 12, creating large numbers that need simplification. Mathematical notation in factorial simplification often involves recognizing patterns. Notice that \(8!\) in both \(12!\) and as the denominator can simplify the expression efficiently

• Factorials \(n!\) are essential in combinatorial principles.
• Operations like division or multiplication of these factorials fit into equations.

By mastering mathematical notation, we can simplify expressions and solve combinatorial problems effectively. Through it, large numeric calculations become more manageable and comprehensible.