Problem 22 - Algebra 2

Question

Evaluate each expression. \(_{8} C_{5}\)

Step-by-Step Solution

Answer

56

Step 1: Understanding Combination Notation

The expression \( _{n} C_{k} \) represents the number of combinations of \( n \) items taken \( k \) at a time without regard to order. It can be calculated using the formula \( _{n} C_{k} = \frac{n!}{k!(n-k)!} \) where \( ! \) denotes factorial.

Step 2: Substitute Values

Substitute \( n = 8 \) and \( k = 5 \) into the formula to get \( _{8} C_{5} = \frac{8!}{5!(8-5)!} \)

Step 3: Calculate Factorials

Evaluate the factorials in the equation. \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) and \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \). For \( (8-5)! \) which is \( 3! = 3 \times 2 \times 1 \)

Step 4: Simplify Expression

Simplify the numerator and denominator by canceling out the common terms. \( \frac{8 \times 7 \times 6 \times 5!}{5! \times 3!} \) simplifies to \( \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \)

Step 5: Perform the Division

Finally, divide the resulting numbers. \( \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \) which is the number of ways to choose 5 items out of 8 without considering the order.

Key Concepts

Combination Notation

In algebra, one of the fundamental concepts you'll encounter is combination notation. This is written as \(_nC_k\) or sometimes \(C(n, k)\), where \(n\) represents the total number of items, and \(k\) is the number of items to choose from that set. Essentially, combination notation answers the question: 'In how many different ways can I select \(k\) items from a set of \(n\) items, ignoring the order in which they are chosen?'
The concept is crucial in probability and statistics when the order of selection does not matter. Imagine you're choosing committee members or creating teams — it is combinations that you would use to figure out the possible groups that can be formed.
The calculation of combinations involves factorial operations, and it's important to understand that it always considers the elements in a set as distinct, even if they appear identical to us. For example, if you have 8 students and want to select 5 to form a team, you'd use combination notation to figure out the possible teams, just like the problem \(_8C_5\).

Factorial

Moving deeper into the realm of combinations, you'll often come across an exclamation mark in math, specifically in expressions like \(n!\). This isn't showing excitement; it denotes a factorial. A factorial is the product of all positive integers from 1 up to that number. In other words, \(n!\) means \(n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1\).
For example, the factorial of 5 (\(5!\)) is \(5 \times 4 \times 3 \times 2 \times 1\), which equates to 120. Factorials are non-negative integers, and by convention, the factorial of zero (\(0!\)) is defined to be 1.
Understanding factorials is essential because they are fundamental to calculating permutations and combinations, which are at the heart of probability. Factorials help us determine the number of ways things can be arranged or selected when the order is (permutations) or is not (combinations) important.

Simplify Mathematical Expressions

A crucial skill in algebra is simplifying mathematical expressions. Simplification can involve basic arithmetic, factoring, combining like terms, or canceling out common factors in fractions. The goal is to reduce an expression to its simplest form, making it easier to work with or understand.
Take our previous example of \(_8C_5\) where we simplified \(\frac{8 \times 7 \times 6 \times 5!}{5! \times 3!}\) into \(\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\). This step involves the cancellation of common terms (in this case, \(5!\)) in the numerator and denominator. It's much like reducing fractions; you’re looking for common factors that can divide both the top and bottom of the fraction to get a simpler, equivalent expression. This process is not only a fundamental aspect of algebraic manipulation, but it also helps to prevent unnecessary calculations by reducing the complexity of the math involved.
When you simplify correctly, you reduce the potential for errors and can more easily see the relationships between numbers and variables, which underpins much of algebraic problem-solving.