Question

Determine the numerical values for the following: a. the number of configurations employing all objects in a six-object set b. the number of configurations employing four objects from a six-object set c. the number of configurations employing no objects from a six-object set d. \(C(50,10)\)

Short answer

The numerical values for the given configurations are as follows: a. 1 configuration using all six objects. b. 15 configurations using four objects from the six-object set. c. 1 configuration using no objects from the six-object set. d. \(C(50,10) = 10,272,278\) ways to choose 10 objects from a set of 50 objects.

Step-by-step solution

Calculate the combinations

Since the set has six objects and we need to find the configurations using all the objects, we need to calculate the number of ways to select 6 objects from 6. Using the combinations formula, we get: \(C(6, 6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!0!}\) Recalling that any number factorial divided by itself is 1, and that 0! is also 1, we have: \(C(6, 6) = \frac{6!}{6!0!} = 1\) Therefore, there is 1 configuration using all the six objects in the set. #b. Configurations using four objects from the six-object set#

Calculate the combinations

We need to find number of ways to select 4 objects out of 6. Using the combinations formula, we get: \(C(6, 4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!}\) Calculating the factorials, we have: \(C(6, 4) = \frac{720}{24\times 2} = \frac{720}{48} = 15\) Therefore, there are 15 configurations using four objects from the six-object set. #c. Configurations using no objects from the six-object set#

Calculate the combinations

For this case, we are not choosing any objects from the set, which means we are selecting zero objects. Using the combinations formula: \(C(6, 0) = \frac{6!}{0!(6-0)!} = \frac{6!}{0!6!}\) Recalling that 0! is 1, we have: \(C(6, 0) = \frac{6!}{0!6!} = 1\) Therefore, there is 1 configuration employing no objects from the six-object set. #d. The value of C(50,10)#

Calculate the combinations

We are asked to find the number of ways to select 10 objects from a set of 50 objects. Using the combinations formula: \(C(50, 10) = \frac{50!}{10!(50-10)!} = \frac{50!}{10!40!}\) We can use a calculator or software to calculate the factorial values and perform the division. The result is: \(C(50, 10) = 10,272,278\) Therefore, there are 10,272,278 ways to choose 10 objects from a set of 50 objects.

Key concepts

Combinations

Combinations are a fundamental concept in combinatorics. When you hear about combinations, think about selecting items from a group. It doesn’t matter what order you pick them in, which makes combinations different from permutations. For example, choosing a team of players from a big group—only the selection matters, not the order in which they were chosen.

The formula for combinations is written as \(C(n, r)\), 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 tells us how many ways we can choose \(r\) items from \(n\) items without caring about the arrangement. Putting this into practice makes solving combinatorics problems a much simpler task.

Factorials

The factorial of a number is a way to multiply a series of descending natural numbers. It's represented by an exclamation mark (!). For example, the factorial of 6 is denoted as \(6!\) and is calculated as \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\).

Factorials are really important in the world of combinations and permutations because they help us calculate how items can be arranged or selected. The small but interesting fact here is that \(0!\) is always equal to 1. This might seem odd, but it forms a fundamental part of combinatorial equations, allowing for expressions involving zero.

Using factorials in equations rearranges large sets of numbers efficiently. Whenever you work with the combination formula \(C(n, r)\), you’ll notice factorials simplify sections of the equation to find the solution.

Combinatorial Problems

Combinatorial problems generally deal with counting and arrangements—a crucial aspect in mathematics that applies to choosing, arranging, and counting objects according to certain rules.

These problems make us think about how to organize objects, such as books on a shelf or teams for a match. In essence, combinatorial math involves solving various selection and arrangement challenges.

They can vary in complexity and might range from straightforward exercises to elaborate puzzles involving multiple choice factors. A key type of combinatorial problem involves using combinations to determine how many ways we can select a certain number of items from a larger set. Training in solving these problems enables better logical reasoning and problem-solving skills.

Mathematical Calculations

Mathematical calculations are essential to solve combinatorics problems, like determining configurations using the combination formula. Performing these calculations involves several steps:

• First, you need to identify all given numbers and understand what is required, whether it's choosing or arranging items.
• Next, apply the necessary formulas, like using factorials in combinations \(\frac{n!}{r!(n-r)!}\), to facilitate calculations.
• Finally, compute the result, carefully performing each step and simplifying wherever possible, especially when dealing with large numbers.

These steps ensure accuracy in finding solutions to problems, like how many ways to select 10 objects from a group of 50, which equals 10,272,278 ways! Handling the math correctly makes challenging problems easy to solve.