Problem 27 - College Algebra

Question

Evaluate each expression. $$ \frac{_{4} C_{2} \cdot_{6} C_{1}}{_{18} C_{3}} $$

Step-by-Step Solution

Answer

The computed value for the expression \(\frac{_4C_2 \cdot _6C_1}{_{18}C_{3}}\) is approximately \(0.0441\). Please note that this is a rounded-off value.

Step 1 - Compute the first combination

Compute \(_4C_2\) using the combination formula: \(_4C_2 = \frac{4!}{2!(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6\)

Step 2 - Compute the second combination

Compute \(_6C_1\) using the combination formula: \(_6C_1 = \frac{6!}{1!(6-1)!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{1 \times 5 \times 4 \times 3 \times 2 \times 1} = 6\)

Step 3 - Compute the denominator

Compute \(_{18}C_{3}\) using the combination formula: \(_{18}C_{3} = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16 \times 15!}{3 \times 2 \times 1 \times 15!} = 816

Step 4 - Evaluate the entire expression

Substitute the values computed in steps 1, 2, and 3 into the given formula: \(\frac{_4C_2 \cdot _6C_1}{_{18}C_{3}} = \frac{6 \times 6}{816} = \frac{1}{22.6667}\)

Key Concepts

Binomial Coefficient

The binomial coefficient is a way to determine the number of ways to choose a subset of objects, irrespective of their order, from a larger set. This concept often comes up in problems dealing with combinations. It is denoted as \( _nC_k \), or sometimes \( C(n, k) \), which reads as "n choose k." Here's the formula for calculating a binomial coefficient:

• \( _nC_k = \frac{n!}{k!(n-k)!} \)

In this formula, \( n! \) (read as "n factorial") means the product of all positive integers up to \( n \), as we'll explore next.
For instance, in the given exercise, computing \( _4C_2 \) involves plugging into the formula, leading to a value that signifies the number of different ways you can choose 2 objects out of a set of 4 objects.

Factorial

Factorial is a key mathematical function used in permutations and combinations. It is expressed with an exclamation mark, such as \( n! \), and represents the product of all positive integers up to \( n \).
For example,

• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \)

Factorials grow very fast as numbers increase, and they are crucial when calculating combinations and permutations. In a combination formula, factorials help determine how to divide the arrangements by accounting for overcounting of indistinct selections where the order doesn't matter.

Probability

Probability is the measure of how likely an event is to occur, expressed as a number between 0 (impossibility) and 1 (certainty). In many problems, probability is closely connected to combinations, as it often represents the ratio of successful outcomes to the total possible outcomes.
A critical component when using combinations in probability is understanding that you often need to compute the number of favorable outcomes divided by all potential outcomes. In the problem given in the exercise, it can be interpreted as identifying how many ways certain groups can be formed compared to all possibilities.

Permutations

Permutations deal with the arrangement of objects where order matters. Unlike combinations, permutations focus on different sequences possible with a set of objects. The formula to calculate permutations, \( _nP_k \), for selecting \( k \) objects from \( n \) is:

• \( _nP_k = \frac{n!}{(n-k)!} \)

This is particularly useful when you need to evaluate how many distinct sequences can be made from a larger set. When solving problems involving both permutations and combinations, recognizing whether the order of selection is crucial helps to decide which formula to employ.
While the original exercise focuses on combinations, understanding permutations enables a fuller grasp of the broader topic of counting methods and their applications in complex scenarios.