Problem 44 - Discrete Mathematics with Applications

Question

Let \(A(n, r)\) denote the number of additions needed to compute \(C(n, r)\) by its recursive definition. Compute each. \(A(5,3)\)

Step-by-Step Solution

Answer

The number of additions needed to compute \(C(5, 3)\) by its recursive definition, \(A(5,3)\), is 7.

Step 1: Write down the recursive formula for the binomial coefficient

The recursive formula for the binomial coefficient is: \[ C(n, r) = C(n-1, r-1) + C(n-1, r) \]

Step 2: Apply the recursive formula for \(C(5, 3)\)

Using the recursive formula, we can write the expression for \(C(5,3)\) as: \[ C(5, 3) = C(4, 2) + C(4, 3) \]

Step 3: Keep applying the recursive formula until base cases are reached

We continue to apply the recursive formula until we reach the base cases, i.e., when \(r = 0\) or when \(n = r\). This step becomes: \[ C(4, 2) = C(3, 1) + C(3, 2) \] \[ C(4, 3) = C(3, 2) + C(3, 3) \] Next, keep applying the recursive formula: \[ C(3, 1) = C(2, 0) + C(2, 1) \] \[ C(3, 2) = C(2, 1) + C(2, 2) \] Finally, all base cases are reached: \[ C(2, 0) = 1\] \[ C(2, 1) = C(1, 0) + C(1, 1) = 1 + 1\] \[ C(2, 2) = 1\]

Step 4: Count the number of additions in the expansion

With the expansion complete, we count the total number of additions required to compute \(C(5,3)\) using its recursive definition: 1 addition from \( C(5, 3) = C(4, 2) + C(4, 3) \) 2 additions from \( C(4, 2) = C(3, 1) + C(3, 2) \) and \( C(4, 3) = C(3, 2) + C(3, 3) \) 3 additions from \( C(3, 1) = C(2, 0) + C(2, 1) \) and \( C(3, 2) = C(2, 1) + C(2, 2) \) 1 addition from \(C(2, 1) = C(1, 0) + C(1, 1) \) The total number of additions required is \(1 + 2 + 3 + 1 = 7\).

Step 5: Final Answer

The number of additions needed to compute \(C(5, 3)\) by its recursive definition, \(A(5,3)\), is 7.

Key Concepts

Recursive Formula

The recursive formula for the binomial coefficient is an essential tool in combinatorics. You can think of it as a mathematical recipe that uses smaller portions of the same problem to solve a larger one. This formula is written as \( C(n, r) = C(n-1, r-1) + C(n-1, r) \), where \(C(n, r)\) represents a binomial coefficient. This expression means that to calculate \(C(n, r)\), you take the sum of \(C(n-1, r-1)\) and \(C(n-1, r)\). This step-by-step approach is vital when dealing with problems involving combinations.

With this recursive method, you break down the problem into manageable chunks by repeatedly applying the formula. The process continues until the simplest base cases are attained, where either \(r = 0\) (which equals 1) or \(n = r\) (also equals 1). This systematic breakdown not only aids in calculations but also helps deepen your understanding of how combinations work.

Combinatorics

Combinatorics is a fascinating area of mathematics focused on counting, arrangement, and combination of objects. The binomial coefficient \( C(n, r) \) plays a pivotal role in combinatorics, representing the number of ways to choose \(r\) elements from \(n\) elements without regard to order. It is often denoted as \(\binom{n}{r}\). When dealing with problems in this field, understanding how to compute these coefficients efficiently is invaluable.

Through the use of recursive formulas, as we see with binomial coefficients, combinatorial problems become easier to tackle. By recognizing patterns and repetitive structures within such problems, we can reduce complex calculations to simpler, repetitive tasks. Using recursion allows for breaking down combinatorial challenges into subsets, leading to more elegant solutions.

Computation Steps

When you compute binomial coefficients recursively, it's crucial to understand and document each step carefully. Starting with \(C(5, 3)\), using the formula \( C(n, r) = C(n-1, r-1) + C(n-1, r) \), begin by applying this formula repeatedly until all possible terms break down into base cases.

Here's a brief outline of the computation steps involved for \( C(5, 3) \):

• First, express it as \( C(5, 3) = C(4, 2) + C(4, 3) \).
• Continue the process by writing \( C(4, 2) = C(3, 1) + C(3, 2) \) and \( C(4, 3) = C(3, 2) + C(3, 3) \).
• Proceed with \( C(3, 1) \) and \( C(3, 2) \) until reaching the simple base cases.

Once reached, these base cases simplify computations significantly. For example, \( C(2, 0) = 1 \), \( C(1, 1) = 1 \), and similarly others. This recursive breakdown requires careful tracking of operations, specifically additions, like this:

• 1 addition for breaking down \( C(5, 3) \).
• 2 additions from \( C(4, 2) \) and \( C(4, 3) \).
• 3 additions for \( C(3, 1) \) and \( C(3, 2) \).
• Finally, 1 more from \( C(2, 1) \).

The total turns out to be 7 additions, showcasing how recursive decomposition helps simplify complex computations.