Question

Evaluate these probabilities: a. \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\) b. \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\) c. \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\)

Short answer

Question: What are the values of the following expressions: a. \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\) b. \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\) c. \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\) Answer: a. The value of the expression \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\) is 0. b. The value of the expression \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\) is 15. c. The value of the expression \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\) is 0.

Step-by-step solution

a. Evaluate \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\)

First, we need to determine the factorials for each combination: - For \(C_1^2\): \(1! = 1\) - For \(C_1^1\): \(1! = 1\) - For \(C_2^3\): \(2! = 2\) Now, we will apply the combination formula for each term: - \(C_1^2 = \frac{1!}{2!(1-2)!} = \frac{1}{2(1)} = 0\) - \(C_1^1 = \frac{1!}{1!(1-1)!} = \frac{1}{1(1)} = 1\) - \(C_2^3 = \frac{2!}{3!(2-3)!} = \frac{2}{6(1)} = \frac{1}{3}\) Finally, we will substitute the values of the combinations into the given expression: \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}} = \frac{0 \times 1}{\frac{1}{3}} = 0\)

b. Evaluate \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\)

First, we need to determine the factorials for each combination: - For \(C_0^4\): \(0! = 1\) - For \(C_2^2\): \(2! = 2\) - For \(C_2^6\): \(2! = 2\) Now, we will apply the combination formula for each term: - \(C_0^4 = \frac{1!}{4!(1-4)!} = \frac{1}{24(1)} = \frac{1}{24}\) - \(C_2^2 = \frac{2!}{2!(2-2)!} = \frac{2}{2(1)} = 1\) - \(C_2^6 = \frac{2!}{6!(2-6)!} = \frac{2}{720(1)} = \frac{1}{360}\) Finally, we will substitute the values of the combinations into the given expression: \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}} = \frac{\frac{1}{24} \times 1}{\frac{1}{360}} = \frac{1}{24} \times 360 = 15\)

c. Evaluate \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\)

First, we need to determine the factorials for each combination: - For \(C_2^2\): \(2! = 2\) - For \(C_1^2\): \(1! = 1\) - For \(C_3^4\): \(3! = 6\) Now, we will apply the combination formula for each term: - \(C_2^2 = \frac{2!}{2!(2-2)!} = \frac{2}{2(1)} = 1\) - \(C_1^2 = \frac{1!}{2!(1-2)!} = \frac{1}{2(1)} = 0\) - \(C_3^4 = \frac{6!}{4!(3-4)!} = \frac{6}{12(1)} = \frac{1}{2}\) Finally, we will substitute the values of the combinations into the given expression: \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}} = \frac{1 \times 0}{\frac{1}{2}} = 0\)

Key concepts

Factorials

Factorials are an essential part of combinatorics because they are used to determine the number of ways items can be arranged or ordered. A factorial, denoted as \( n! \), is the product of all positive integers up to a specified number \( n \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

Factorials are especially important in calculating permutations and combinations. Here's why:

• The factorial function helps in determining the likelihood of events by defining the number of possible arrangements or sequences.
• Factorials are used to eliminate the repetition of permutations when calculating combinations where the sequence does not matter.
• In probability problems, understanding factorials can simplify the calculation of complex expressions.

Be mindful that for zero factorial, the value is always defined as \( 0! = 1 \). This is a unique property that simplifies many equations in combinatorial mathematics.

Probability Evaluation

Evaluating probabilities often involves understanding the relationships between different combinations and permutations. In the context of probability, combinatorics helps us to count and assess the likelihood of different outcomes.

When evaluating probabilities:

• Start by defining the total number of possible outcomes using combinations or permutations.
• Identify the number of successful outcomes that match the criteria of the event.
• Express the probability as a ratio or a fraction of successful outcomes over the total number of possible outcomes.

The use of factorials and combinations as seen in the provided expressions involves breaking down complex ratios into simpler components. This provides a straightforward path to evaluate which options are more likely to occur. In our original exercise example, we see probability evaluation through dividing combinations to find ratios of likelihood in different scenarios.

Combination Formula

The combination formula is a tool we use to calculate how many ways we can choose items without regard to order from a larger set. It is denoted as \( C(n, r) \) or \( \binom{n}{r} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose.

The formula is: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] This formula calculates how many different groups of `r` items can be formed from a set of `n` items.

Key points:

• Unlike permutations, combinations do not consider the order of items. Only the selection matters.
• The combination formula utilizes factorials to divide the total arrangements by the duplications that arise from internal ordering of selected items.
• This formula is fundamental in areas like binomial probability, lottery calculations, and any scenario involving selection without order.

In our exercise, we apply the combination formula to encapsulate both the concept of choosing items and the likelihood that sequence and order play no role in the outcome, showing its practical importance.