/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Evaluate each number. $$ C(1... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each number. $$ C(100,98) $$

Short Answer

Expert verified
The value of \(C(100, 98)\) is 4950.

Step by step solution

01

In the expression \(C(100, 98)\), n = 100 and k = 98. We want to select 98 items from a set of 100 items. #Step 2: Apply the combination formula to find the number of combinations #

To evaluate \(C(100, 98)\), we will use the formula: \[C(n, k) = \frac{n!}{k!(n-k)!}\] Plugging in the values of n and k, we get: \[C(100, 98) = \frac{100!}{98!(100-98)!}\] #Step 3: Simplify the expression#
02

Now, let's simplify the expression: \[C(100, 98) = \frac{100!}{98!(2)!}\] Since we know that \(2! = 2 * 1 = 2\), we can further simplify the expression: \[C(100, 98) = \frac{100!}{98! * 2}\] #Step 4: Calculate the factorial values, and find the result#

To find the value of \(100!\) and \(98!\), we can use the property of factorials that says: \[(n-1)! = \frac{n!}{n}\] Applying this property to our expression: \[\frac{100!}{98! * 2} = \frac{(100)(99)(98!)}{98!(2)}\] Now, we can cancel out \(98!\) from both the numerator and the denominator: \[= \frac{100 * 99}{2}\] Now, calculate the result: \[= \frac{100 * 99}{2} = 50 * 99 = 4950\] So, the value of \(C(100, 98)\) is 4950.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a field of mathematics concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is closely related to many other areas of mathematics and has many practical applications in science, technology, and computer science. The counting comes in many flavors: there are permutations, combinations, partitions, and more. For example, if you want to figure out how many different ways you can pick a team of 3 from a group of 10, combinatorics is the area of math you'd consult. Combinatorics can get quite sophisticated and often involves additional principles or mathematical concepts.
Factorial Notation
Factorial notation is used to simplify the expression of permutations and combinations. The factorial of a number, represented as \(n!\), is the product of all positive integers up to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). It's important to remember that \(0!\) is defined to be 1. Factorials grow very quickly with the value of \(n\), which is why they appear in counting problems where the number of possibilities explodes with the addition of each new item.

For combinatorial calculations, factorials are the building blocks. Understanding how to manipulate and simplify expressions using factorial notation is crucial for solving combinations and permutations.
Permutations and Combinations
Permutations and combinations are fundamental concepts in combinatorics that describe two different ways of selecting items from a group. Permutations are interested in the order of selection, while combinations are not.

A permutation is an ordered arrangement of objects. For instance, if we're forming teams and the order matters (like a batting lineup in baseball), we'd use permutations. Combinations, on the other hand, are used when the order doesn't matter, such as picking members for a committee. In other words, combinations are for when you care about 'who is on the team,' and permutations are for when you care about 'who is where on the team.'

To differentiate, the number of ways to arrange \(k\) items out of \(n\) is given by permutations, while combinations tell us how many ways we can choose \(k\) items from a set of \(n\), irrespective of their order.
Evaluating Combinations
Evaluating combinations involves using the combination formula. The general form of the combination formula is \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \(n\) is the total number of items, and \(k\) is the number of items to be chosen. This formula calculates the different possible combinations of \(k\) items from a set of \(n\) items.

When simplifying combination expressions, there's often a lot of cancellation that happens, which makes manual calculation doable. For example, in the expression \(C(100, 98)\), the large factorials cancel each other out, leaving a much simpler multiplication and division.

Using factorial properties effectively, as shown in the original example, helps in 'reducing' what looks like an intimidating calculation to a few simple steps. This is a very handy approach especially when dealing with large numbers often encountered in combinatorics, saving time and minimizing the potential for errors.

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