Problem 21 - Algebra 2 and Trigonometry

Question

In \(3-22,\) evaluate each expression. $$ _{8} C_{3} \div_{8} C_{5} $$

Step-by-Step Solution

Answer

The expression evaluates to 1.

Step 1: Understand the Notation

The expression involves combinations denoted as \(_nC_k\) which represents the number of ways to choose \(k\) items from \(n\) items without regard to the order of selection. The formula for this is given by: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(!\) denotes factorial.

Step 2: Compute \(_8C_3\)

Substitute \(n = 8\) and \(k = 3\) into the combination formula:\[ _8C_3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} \]Simplify this by canceling terms:\[ 8 \cdot 7 \cdot 6 = 336 \] and\[ 3! = 3 \cdot 2 \cdot 1 = 6 \]So,\[ _8C_3 = \frac{336}{6} = 56 \]

Step 3: Compute \(_8C_5\)

Using the identity \(_nC_k = _nC_{n-k}\), note that \(_8C_5 = _8C_3\), which simplifies our work since we've already computed \(_8C_3 = 56\). Thus, \(_8C_5 = 56\).

Step 4: Evaluate the Expression

The expression to evaluate is \(_8C_3 \div _8C_5\). Substitute the values found in the previous steps:\[ \frac{_8C_3}{_8C_5} = \frac{56}{56} = 1 \]

Key Concepts

Factorial

The concept of a factorial, denoted by the symbol \(!\), is vital for understanding combinations and permutations in mathematics. A factorial is the product of all positive integers less than or equal to a given positive integer \(n\). We express this as \(n!\). For example, \(5!\) means \(5 \times 4 \times 3 \times 2 \times 1 = 120\).

The factorial function grows very quickly with larger numbers, making it crucial for calculating large possibilities in combinatorial problems. Here's a tip: remember that \(0! = 1\) by definition, which is sometimes surprising but simplifies many expressions and calculations in mathematics

Factoring the expression efficiently uses the idea of cancellation, like in the exercise above, where the factorial divisions cancel out unnecessary terms, simplifying calculations.

Binomial Coefficient

The binomial coefficient is key to solving problems involving combinations. It tells us how many ways we can choose \(k\) items from \(n\) items, where order does not matter. This is represented by the notation \(_nC_k\) or \(\binom{n}{k}\).

Calculating \(\binom{n}{k}\) involves factorials, using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

This formula arranges the items into all possible ways, then divides out the repetitions (since the order is not important), giving the correct count of unique combinations. For example, \(_8C_3\) means choosing 3 items out of 8, calculated as \(\frac{8!}{3!5!}\). This was simplified by recognizing that \( \binom{n}{k} = \binom{n}{n-k} \), which helped in the original problem to quickly find that \(_8C_5\) is equal to \(_8C_3\).

This symmetry simplifies many problems and is a useful trick for efficient calculations.

Combinatorics

Combinatorics is the branch of mathematics focused on counting, arranging, and finding patterns. It deeply involves working with permutations and combinations. In our exercise, we worked with combinations, which are selections where order doesn't matter.

Combinatorics often involves strategies to break complex counting problems into simpler ones, using principles like combinations, permutations, and sometimes more advanced topics like the pigeonhole principle or inclusion-exclusion principle.

Understanding these basics aids in solving more complex real-world problems, whether you're compiling statistics, organizing data, or dealing with probability. The foundation is important; starting from understanding how to calculate combinations helps in building toward solving intricate counting problems effectively.