Problem 27 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The value of the expression is approximately 0.04412.

Step 1: Calculate Numerator Part 1 - _4C_2

First, calculate the first part of the numerator, which is _4C_2 (four choose two). Apply the formula \( _nC_r = \frac{n!}{r!(n-r)!} \) so \( _4C_2 = \frac{4!}{2!(4-2)!} = \frac{24}{2*2} = 6.\)

Step 2: Calculate Numerator Part 2 - _6C_1

Next, calculate the second part of the numerator, which is _6C_1 (six choose one). Apply the formula \( _nC_r = \frac{n!}{r!(n-r)!} \) so \( _6C_1 = \frac{6!}{1!(6-1)!} = \frac{720}{5*4*3*2*1} = 6.\)

Step 3: Calculate Denominator - _18C_3

Then, calculate the denominator, which is _18C_3 (eighteen choose three). Apply the formula \( _nC_r = \frac{n!}{r!(n-r)!} \) so \( _18C_3 = \frac{18!}{3!(18-3)!} = \frac{6,402,373,705,728,000}{6*5*4*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1} = 816.\)

Step 4: Final Calculation

Finally, evaluate the full expression by multiplying the two results from the numerator (6 * 6 = 36) and dividing by the result from the denominator (36 / 816). Simplify to get the final result, approximately 0.04412.

Key Concepts

Understanding the Binomial Coefficient

In combinatorics, the binomial coefficient is a fundamental concept that helps us determine the number of combinations of a given set of items. It is denoted often as \( _nC_r \) or \( \binom{n}{r} \), where \( n \) represents the total number of items and \( r \) represents the number of items to choose. To compute this, you use the formula:

• \( _nC_r = \frac{n!}{r!(n-r)!} \)

Here, \( ! \) is the factorial operation, and it means you multiply all whole numbers from the chosen number down to 1. This formula is quite handy when you are trying to figure out how many ways you can choose a subset of \( r \) items out of \( n \) total items without regard to order.
For instance, "four choose two" which translates to \( _4C_2 \), means choosing 2 items out of 4. Using our formula:

• Calculate \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• Calculate \( 2! = 2 \times 1 = 2 \)
• \( (4-2)! = 2! = 2 \)
• Put it all together: \( _4C_2 = \frac{24}{2 \times 2} = 6 \)

Thus, there are 6 different ways to choose 2 items from 4.

Exploring Factorials

Factorial is an essential concept in combinatorics, closely tied with calculations involving binomial coefficients. A factorial, denoted by the symbol \( ! \), represents the product of an integer and all the integers below it down to 1. It is defined only for non-negative integers. For example:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 1! = 1 \)
• \( 0! = 1 \) by definition

Applying factorials allows us to express combinations and permutations in a manageable way. In our exercise, factorials were vital to compute the values of \( _nC_r \). The factorial of a number helps determine how arrangements can be made. For example, the calculation \( 18! \) in \( _18C_3 \) means calculating a vast succession of numbers, making factorials pivotal for such large calculations in probability and combinatorics.
It's important to become familiar with factorials, as these are building blocks for understanding more complex probability scenarios.

Probability Calculation and Applications

Probability calculations are a crucial application of combinatorics, crucial for understanding real-world situations where outcomes are uncertain. Probability offers a method to predict the likelihood of different possible outcomes. It can be calculated by using combinations when it comes to choosing specific outcomes from a larger set.
This involves dividing the number of successful outcomes by the total possible outcomes, aligning well with the binomial coefficient calculation.

• Formula for probability: \( \text{Probability} = \frac{\text{Number of Successful Outcomes}}{\text{Total Possible Outcomes}} \)

In the exercise, the calculation involved determining success by using the binomial coefficient. First, we determine the possible combinations (\( _4C_2 \) and \( _6C_1 \)) and calculate their product. Then, determine the total number of configurations (\( _18C_3 \)). The probability calculation then divides these to find the likelihood of a particular configuration of choices, yielding:

• \( \frac{6 \times 6}{816} \approx 0.04412 \)

Therefore, understanding this process is invaluable in fields like statistics, risk assessment, and decision-making where predicting different outcomes is essential.