Problem 13
Question
In Exercises 5 - 14, calculate the binomial coefficient. \( \dbinom{100}{98} \)
Step-by-Step Solution
Verified Answer
The binomial coefficient \(\dbinom{100}{98}\) is 4950.
1Step 1: Apply the binomial coefficient formula
We can calculate the binomial coefficient using its formula: \(\dbinom{100}{98} = \frac{100!}{98! * (100-98)!}\). This simplifies to \(\dbinom{100}{98} = \frac{100!}{98! * 2!}\)
2Step 2: Simplify the factorial terms
Next, consider that \(n!\) is equal to \(n * (n-1)!\). Thus, \(100!\) can be written as \(100 * 99 * 98!\) and \(2!\) is simply \(2\). Now the expression becomes \(\dbinom{100}{98} = \frac{100 * 99 * 98!}{98! * 2}\). Notice that \(98!\) cancels out in the numerator and the denominator.
3Step 3: Compute the binomial coefficient
Now we are left with a simple multiplication and division: \(\dbinom{100}{98} = \frac{100 * 99}{2}\). Performing the calculation yields \(\dbinom{100}{98} = 4950\).
Key Concepts
Factorial ConceptsCombinatoricsCalculation Methods
Factorial Concepts
Factorials are a fundamental concept in mathematics, especially in permutations and combinations. The factorial of a non-negative integer, denoted by an exclamation mark (!), represents the product of all positive integers up to that number. For instance, the factorial of 5 is written as \(5!\), and calculated as \(5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials grow very quickly as numbers increase, which makes them crucial in calculations involving large sets, such as binomial coefficients.
An important property of factorials is that \(n! = n \times (n-1)!\). This recursive relationship is often used to simplify calculations, like when simplifying the expression \(100! = 100 \times 99 \times 98!\). By mastering factorials, you gain insight into the backbone of various combinatorial calculations.
Factorials grow very quickly as numbers increase, which makes them crucial in calculations involving large sets, such as binomial coefficients.
An important property of factorials is that \(n! = n \times (n-1)!\). This recursive relationship is often used to simplify calculations, like when simplifying the expression \(100! = 100 \times 99 \times 98!\). By mastering factorials, you gain insight into the backbone of various combinatorial calculations.
Combinatorics
Combinatorics is the branch of mathematics dealing with combinations of objects. It plays a crucial role in calculating probabilities, arrangements, and combinations efficiently. A key concept is the binomial coefficient, often denoted as \(\binom{n}{r}\).
The binomial coefficient \(\binom{n}{r}\) represents the number of ways to select \(r\) objects from a set of \(n\) objects, without concern for the order of selection. It is calculated using the formula:
Understanding combinatorics is essential for solving many types of math problems, from probability theory to algebra. Mastery involves being proficient with concepts such as factorial and the use of formulas like the binomial coefficient.
The binomial coefficient \(\binom{n}{r}\) represents the number of ways to select \(r\) objects from a set of \(n\) objects, without concern for the order of selection. It is calculated using the formula:
- \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)
Understanding combinatorics is essential for solving many types of math problems, from probability theory to algebra. Mastery involves being proficient with concepts such as factorial and the use of formulas like the binomial coefficient.
Calculation Methods
When approaching calculations involving factorials and binomial coefficients, it is essential to simplify your expressions. One way to do this is by recognizing patterns or properties within the factorial expressions.
For example, in computing \(\binom{100}{98}\), notice that \(100!\) can be expanded to \(100 \times 99 \times 98!\). The \(98!\) terms in the numerator and denominator cancel each other out, simplifying the expression to \(\frac{100 \times 99}{2}\).
Using this simplification reduces the calculation's complexity, making it easier and quicker to find solutions. Breaking down problems into smaller, simpler components is a common strategy in mathematics, particularly in combinatorics and probability.
To further ease calculations, making mental notes of smaller factorial values (like \(2! = 2\) or \(3! = 6\)) is helpful, allowing quicker simplification and computation.
For example, in computing \(\binom{100}{98}\), notice that \(100!\) can be expanded to \(100 \times 99 \times 98!\). The \(98!\) terms in the numerator and denominator cancel each other out, simplifying the expression to \(\frac{100 \times 99}{2}\).
Using this simplification reduces the calculation's complexity, making it easier and quicker to find solutions. Breaking down problems into smaller, simpler components is a common strategy in mathematics, particularly in combinatorics and probability.
To further ease calculations, making mental notes of smaller factorial values (like \(2! = 2\) or \(3! = 6\)) is helpful, allowing quicker simplification and computation.
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Problem 13
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