Problem 12 - College Algebra

Question

For the following exercises, evaluate the binomial coefficient. $$ \left(\begin{array}{l} 200 \\ 199 \end{array}\right) $$

Step-by-Step Solution

Answer

The binomial coefficient \( \binom{200}{199} \) is 200.

Step 1: Understanding the Binomial Coefficient

The binomial coefficient, denoted as \(\binom{n}{k}\), represents the number of ways to choose \(k\) items out of \(n\) without regard to order. For this problem, we have \(n = 200\) and \(k = 199\).

Step 2: Applying the Binomial Coefficient Formula

The formula for the binomial coefficient is given by \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). For \(\binom{200}{199}\), we need to calculate \(\frac{200!}{199!(200-199)!}\).

Step 3: Simplifying the Expression

Simplify the expression \(\frac{200!}{199!(1)!}\). Since \((1)! = 1\), this simplifies to \(\frac{200!}{199!}\).

Step 4: Further Simplifying by Canceling the Factorials

When dividing \(200!\) by \(199!\), all the numbers from \(1\) to \(199\) cancel out, leaving us with \(200\). So, \(\frac{200!}{199!} = 200\).

Key Concepts

Factorials

A factorial, denoted by an exclamation mark, is a way to represent the product of all positive integers up to a given number. For any positive integer \(n\), the factorial of \(n\), written as \(n!\), is calculated as follows:

• Example: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
• By definition, \(0! = 1\)

Factorials are used in many mathematical concepts, including permutations, combinations, and the calculation of binomial coefficients. They provide a convenient way to describe quantities that grow very quickly with increasing \(n\).
In the context of the problem, we calculated \(200!\) and \(199!\) to find the binomial coefficient \(\binom{200}{199}\). For this coefficient, the expression \(200! / 199!\) simplifies quite easily because all factors from \(1\) to \(199\) cancel each other out, leaving the factorial calculations very manageable.

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and selection of objects. It plays a significant role in various fields, like computer science, cryptography, and pure mathematics. The primary goals in combinatorics involve finding efficient ways to count and choose groups of items.
In relation to the binomial coefficient, combinatorics helps us determine the number of ways to choose \(k\) items from a set of \(n\) items, where the order of selection does not matter. This is precisely the problem tackled in our exercise, where we looked at choosing \(199\) items from \(200\).

• This is visually depicted using Pascal's Triangle or calculated using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

The power and utility of combinatorics stem from its systematic approach to solve problems that would be too vast or complex by simple enumeration.

Permutation and Combination

Permutation and combination are core concepts in combinatorics that deal with arranging and selecting items.

• Permutations: Concerned with the arrangement of items in a specific order. If order matters, then the problem is about permutations. For \(n\) items, the number of permutations when selecting \(k\) is given by \(\frac{n!}{(n-k)!}\).

• Example: Arranging three books on a shelf. Order matters, so you want to find permutation.

• Combinations: Unlike permutations, combinations are about selecting items without regard to order. This is what the exercise focused on. The binomial coefficient is a method for finding combinations.

• Example: Choosing 2 fruits from a basket of 5. The order doesn't matter, so combinations are what you seek.

Understanding the difference between permutations and combinations is essential as it determines the approach to solving a particular problem. Our exercise involving \(\binom{200}{199}\) is an example of combinations, where order is irrelevant.