Problem 19 - Precalculus: Mathematics for Calculus

Question

Evaluate the expression. $$\left(\begin{array}{c}100 \\ 98\end{array}\right)$$

Step-by-Step Solution

Answer

4950

Step 1: Understand the Binomial Coefficient

The expression \( \binom{100}{98} \) represents a binomial coefficient, which is used to find the number of ways to choose 98 elements from a set of 100 elements. This is mathematically expressed as \( \frac{100!}{98!(100-98)!} \).

Step 2: Simplify the Factorial Expression

We know that \( 100! = 100 \times 99 \times 98! \). Inserting this into our formula gives us \( \frac{100 \times 99 \times 98!}{98! \times 2!} \).

Step 3: Cancel Out the Common Factorials

Now, cancel \( 98! \) in both the numerator and the denominator to simplify the expression to \( \frac{100 \times 99}{2!} \).

Step 4: Calculate 2!

Calculate \( 2! = 2 \times 1 = 2 \).

Step 5: Final Calculation

Substitute the value of \( 2! \) into the equation to get \( \frac{100 \times 99}{2} = \frac{9900}{2} = 4950 \). This is the value of \( \binom{100}{98} \).

Key Concepts

Binomial Coefficient

In the realm of combinatorics, the binomial coefficient is a fundamental concept that represents the number of ways to choose a subset of elements from a larger set. Expressed as \( \binom{n}{k} \), it calculates the number of ways to select \( k \) elements from \( n \) distinct items. This is particularly useful in problems involving combinations and is often read as "n choose k."

The formula for the binomial coefficient is:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

This equation considers the total number of arrangements \( n! \) divided by the arrangements of the chosen subset \( k! \) and the non-chosen subset \( (n-k)! \).

For instance, in our example where you need to find \( \binom{100}{98} \), it can be simplified as "100 choose 98," and it's equivalent to choosing 2 elements from 100 because choosing 98 out of 100 is the same as leaving out exactly 2 from the total 100 elements.

Factorial Notation

Factorial notation is a way of expressing a series of descending multiplications. Here's how it works: the factorial of a positive integer \( n \), denoted by \( n! \), is the product of all positive integers less than or equal to \( n \).

For example, \( 5! \) is calculated as:

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

This concept simplifies calculations in the realm of permutations and combinations by providing a structured way of calculating large numbers of arrangements.

In the context of the binomial coefficient \( \binom{n}{k} \), factorial notation helps in breaking down complex calculation processes. In our exercise, \( 100! \) refers to multiplying all numbers from 100 down to 1. However, this is simplified, since in calculations like \( \binom{100}{98} \), many terms cancel out, leaving smaller, more manageable calculations, such as \( 100 \times 99 \).

Permutations and Combinations

Permutations and combinations are two approaches to arranging and selecting items from a set. They are cornerstones in combinatorics, helping us calculate possibilities in complex scenarios.

**Permutations** deal with the arrangement of items where order matters. For example, if you want to arrange 3 books on a shelf, you would calculate this using permutations, as the order (or permutation) of books matters.

**Combinations**, on the other hand, focus on selection where the arrangement doesn't matter. Binomial coefficients are all about combinations because they count how many ways you can select items from a group, regardless of order.

In the given exercise, we used combinations, specifically, to determine how many ways we can choose 98 items from a set of 100. This doesn't involve arranging those 98 items, just selecting them. This distinct difference between permutations and combinations is essential for solving problems accurately in combinatorics.