Question

Prove that \(C(n, r)>C(n, r-1)\) if \(r

Short answer

For \( r < \frac{n}{2} \), \( C(n, r) > C(n, r-1) \) holds true in Pascal's triangle.

Step-by-step solution

Understand the binomial coefficient

The binomial coefficient is defined as \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \(!\) denotes factorial. This represents the number of ways to choose \(r\) elements from a set of \(n\) elements.

Set up the inequality

We need to prove that \( C(n, r) > C(n, r-1) \) for \( r < \frac{n}{2} \). We start by comparing the two expressions: \( \frac{n!}{r!(n-r)!} > \frac{n!}{(r-1)!(n-r+1)!} \).

Simplify the inequality

Simplify the comparative expression: \( \frac{1}{r!(n-r)!} > \frac{1}{(r-1)!(n-r+1)!} \). This becomes \( \frac{1}{r(n-r)} > \frac{1}{(n-r+1)} \) by canceling and rearranging terms.

Further simplify

Multiply both sides by \((r-1)!(n-r+1)!\) to eliminate factorials: \( (n-r+1) > r(n-r) \).

Solve the inequality

The inequality \( n-r+1 > r(n-r) \) simplifies further to \( n-r+1 > rn-r^2 \), leading us to \( n+1 > rn-r^2+r \). Simplifying, we get \( n+1 > r(n-r) \).

Check condition for r

We know \( n > 2r \) or \( r < \frac{n}{2} \) holds and ensures our simplification maintains a true inequality, satisfying the given condition for \(C(n,r) > C(n, r-1)\).

Key concepts

Binomial Coefficient

The binomial coefficient is a fundamental concept in combinatorics, representing the number of ways to choose a certain number of elements from a larger set. It is denoted as \( C(n, r) \) and calculated using the formula: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] where \(!\) signifies factorial. The factorial numbers multiply all whole numbers from the specified number down to one, thus \( n! \) means \( n \times (n-1) \times (n-2) \times \ldots \times 1 \).

By understanding binomial coefficients, we learn how they reflect arrangements and selections within sets. For example, this coefficient counts the different ways to select \( r \) items from a larger group of \( n \) items without regard to order. It's an essential building block for later understanding in more complex mathematics, such as probabilities and statistics.

Factorial

Factorials are key to computing binomial coefficients. The notation \( n! \) means the product of all positive integers up to \( n \). So, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). This operation grows rapidly, making factorials very large, very fast.

Factorials have many uses. They calculate permutations – all possible arrangements of a set. For instance, if you want to know how to arrange \( 4 \) books on a shelf, the answer is \( 4! \), which is 24 ways. In our problem, factorials help determine binomial coefficients, which in turn find solutions to combinatorial questions like how many ways to choose items from larger collections without replacement.

Inequality in Binomial Coefficients

We encounter inequalities between binomial coefficients often when comparing different combinations. Specifically, in the context of Pascal's Triangle, if you want to show one coefficient is greater than another, inequalities are useful.

In the exercise, we prove that \( C(n, r) > C(n, r-1) \) when \( r < \frac{n}{2} \). Comparing the coefficients involves setting up and simplifying inequalities because the arrangement of elements changes as \( r \) approaches \( \frac{n}{2} \). By manipulating the factorial expressions, the inequality \( n+1 > r(n-r) \) crystalizes, confirming this relationship. This type of inequality analysis can extend to many other mathematical proofs and be powerful in theoretical and applied mathematics.

Combinatorics

The field of combinatorics explores how we count and arrange objects. It provides tools to understand various forms of arrangement and selection, especially useful in complex situations. Binomial coefficients and factorials are among its best-known concepts.

In combinatorics, problems often involve figuring out the number of ways to choose or arrange objects, whether it's selecting teams, forming committees, or arranging sequences. The study forms the backbone of discrete mathematics, linking to probability, set theory, and algebra.

With situations as in the exercise, combinatorics dives deep into counting the different combinations where certain conditions (like \( r < \frac{n}{2} \)) are applicable. This understanding not only helps solve mathematical problems but also applies to computer science, finance, and game theory, making combinatorics a rich and fascinating study.