Problem 48 - College Algebra

Question

Find \(\left(\begin{array}{c}{n} \\\ {k-1}\end{array}\right)+\left(\begin{array}{c}{n} \\ {k}\end{array}\right)\) and write the answer as a binomial coefficient in the form \(\left(\begin{array}{l}{n} \\ {k}\end{array}\right)\) Prove it. Hint Use the fact that, for any integer \(p,\) such that \(p \geq 1\) \(p !=p(p-1) !\).

Step-by-Step Solution

Answer

\(\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}\).

Step 1: Understanding Binomial Coefficients

The expression \(\binom{n}{k}\) is defined as \(\frac{n!}{k!(n-k)!}\). We need to find \(\binom{n}{k-1} + \binom{n}{k}\).

Step 2: Writing Down the Expression

We'll express both terms using factorials: \[\binom{n}{k-1} = \frac{n!}{(k-1)!(n-k+1)!}\] \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] We will sum these two expressions.

Step 3: Finding a Common Denominator

The denominators are \((k-1)!(n-k+1)!\) and \(k!(n-k)!\). The lowest common denominator (LCD) is \(k!(n-k+1)!\).

Step 4: Rewriting the Terms with a Common Denominator

Rewrite the first term to have the common denominator:\[\frac{n!}{(k-1)!(n-k+1)!} \times \frac{k}{k} = \frac{k \cdot n!}{k!(n-k+1)!}\]Rewrite the second term:\[\frac{n!}{k!(n-k)!} \times \frac{n-k+1}{n-k+1} = \frac{(n-k+1) \cdot n!}{k!(n-k+1)!}\]

Step 5: Adding the Terms

Now add the two terms:\[\frac{k \cdot n!}{k!(n-k+1)!} + \frac{(n-k+1) \cdot n!}{k!(n-k+1)!} = \frac{(k + n-k+1) n!}{k!(n-k+1)!}\]Simplify the numerator:\[k + n - k + 1 = n + 1\]Thus, the expression becomes:\[\frac{(n+1) n!}{k!(n-k+1)!}\]

Step 6: Simplifying the Expression

Notice that the previous expression corresponds to the definition of a binomial coefficient:\[\frac{(n+1)!}{k!(n+1-k)!}\], which is equivalent to \(\binom{n+1}{k}\).

Step 7: Conclusion

Therefore, \(\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}\). This isn't just an algebraic trick; it reflects the combinatorial principle of adding successively larger subsets.

Key Concepts

Factorial

The factorial, denoted by an exclamation mark (!), is a concept that finds great utility in mathematics, particularly in combinatorics and probability. The factorial of a non-negative integer, say \(n\), is symbolized as \(n!\) and represents the product of all positive integers from 1 to \(n\). This means:

• \(n! = n \times (n-1) \times (n-2) \times \ldots \times 1\)

For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\). Factorials are crucial when dealing with permutations and combinations, as they help in calculating the total number of ways to arrange or select items. In the context of binomial coefficients, factorials are used to simplify expressions, making calculations more manageable. For instance, to define the binomial coefficient \(\binom{n}{k}\), which is used in the exercise above, you would use the formula \(\frac{n!}{k!(n-k)!}\), encapsulating the factorial’s role perfectly.

Combinatorial Principle

The combinatorial principle is an insightful tool for understanding and proving identities involving binomial coefficients. Essentially, it provides a framework for counting the number of ways to choose subsets or arrange objects. The expression \(\binom{n}{k}\) is a binomial coefficient that represents the number of combinations of \(n\) items taken \(k\) at a time. In the exercise above, we are tasked to find \(\binom{n}{k-1} + \binom{n}{k}\) and express it as another binomial coefficient. This equates to combining subsets of sizes \(k-1\) and \(k\) from a set of \(n\), which by the combinatorial principle can be represented as \(\binom{n+1}{k}\). This is not just an algebraic manipulation but a logical deduction indicating how different selections overlap to form a larger set, demonstrating the elegance of combinatorial thinking.

Common Denominator

Finding a common denominator is a fundamental skill in algebra necessary for adding fractions. In the context of our exercise, it allows us to add the two binomial expressions effectively:

• \(\binom{n}{k-1} = \frac{n!}{(k-1)!(n-k+1)!}\)
• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

To combine these, we identify the lowest common denominator (LCD), which is \(k!(n-k+1)!\) in this case. This process aligns the terms so they can be directly added together, crucial for algebraic operations that merge fractions into simpler, unified expressions. It's an important aspect of simplifying expressions that appear complex at first glance.

Algebraic Simplification

Algebraic simplification is all about reducing expressions to their simplest forms, making them easier to work with and understand. In the solution to this exercise, it involves smartly handling the components of binomial expressions to reveal their underlying simplicity. After identifying a common denominator, the fractions are altered (via equivalent multipliers) to have uniform denominators. This permits effective addition.The expression \(\frac{(k + n-k+1) n!}{k!(n-k+1)!}\) simplifies in the numerator to \((n+1)\), showcasing the magic of simplification:

• Original numerator: \(k + n - k + 1 = n + 1\)

Finally, this reduces to \(\frac{(n+1)!}{k!(n+1-k)!}\), succinctly expressing it as a single binomial coefficient \(\binom{n+1}{k}\). This highlights how algebraic simplification not only streamlines our solutions but also underlines the interplay between arithmetic operations and logical reasoning.