Question

Consider the following problem: How many ways can a committee of four people be selected from a group of 10 people? There are many other similar problems in which you are asked to find the number of ways to select a set of items from a given set of items. The general problem can be stated as follows: Find the number of ways \(r\) different things can be chosen from a set of \(n\) items, in which \(r\) and \(n\) are nonnegative integers and \(r \leq n .\) Suppose \(C(n, r)\) denotes the number of ways \(r\) different things can be chosen from a set of \(n\) items. Then, \(C(n, r)\) is given by the following formula: \(C(n, r)=\frac{n !}{r !(n-r) !}\) in which the exclamation point denotes the factorial function. Moreover, \(C(n, 0)=C(n, n)=1 .\) It is also known that \(C(n, r)=C(n-1, r-1)+C(n-1, r)\) a. Write a recursive algorithm to determine \(C(n, r)\). Identify the base case \((\mathrm{s})\) and the general case \((\mathrm{s})\) b. Using your recursive algorithm, determine \(C(5,3)\) and \(C(9,4)\)

Short answer

C(5, 3) = 10 and C(9, 4) = 126 using the recursive formula.

Step-by-step solution

Understanding the Problem

We need to find the number of ways to select a subset of 4 people from a group of 10, which is a type of combination problem. For this, combinations are used, denoted as \( C(n, r) \).

Identify Combinatorial Formula

The formula for combinations is \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \(!\) denotes a factorial.

Define Recursive Algorithm for C(n, r)

The recursive relationship for combinations is given by:- **Base case:** \( C(n, 0) = C(n, n) = 1 \).- **Recursive case:** \( C(n, r) = C(n-1, r-1) + C(n-1, r) \).

Apply Recursive Algorithm to C(5,3)

Using the recursive relationship, we calculate \( C(5, 3) \):- \( C(5, 3) = C(4, 2) + C(4, 3) \)- \( C(4, 2) = C(3, 1) + C(3, 2) \)- \( C(4, 3) = C(3, 2) + C(3, 3) \)- From the base cases: \( C(3, 1) = 3 \), \( C(3, 2) = 3 \), and \( C(3, 3) = 1 \)- Substitute back: \( C(4, 2) = 3 + 3 = 6 \), \( C(4, 3) = 3 + 1 = 4 \)- Finally: \( C(5, 3) = 6 + 4 = 10 \)

Apply Recursive Algorithm to C(9,4)

Calculate \( C(9, 4) \) using the recursive formula:- Simplifying using the recursive steps:1. \( C(9, 4) = C(8, 3) + C(8, 4) \)2. Break into smaller problems until base cases are reached: - Calculate values like \( C(8, 3), C(8, 4) \) similar to step 43. Use previously determined values to calculate up to \( C(9, 4) \)4. Final calculations give: \( C(9, 4) = 126 \)

Key concepts

Combinations: Picking with Precision.

Understanding combinations is similar to picking a small team from a larger group. Imagine you have a group of 10 people and you want to form a committee of 4. Each possible selection is a combination. We represent combinations with the notation \( C(n, r) \), where \( n \) is the total number of items (or people) to choose from, and \( r \) is the number of items to choose.

The formula to calculate combinations is:

• \( C(n, r) = \frac{n!}{r!(n-r)!} \)

Here, the factorial function (denoted by \(!\)) plays a key role. This formula ensures that the order of selection doesn't matter - only the chosen members matter in combinations.

Combinations focus mainly on selection, not arrangement. It's like building teams, where who you pick is the only thing that matters.

Factorial: Multiply to Zero!

Factorials might sound complex, but they essentially denote repetitive multiplication. For any integer \( n \), the factorial is written as \( n! \) and calculated as \( n \times (n-1) \times (n-2) \times \, ...\, \times 2 \times 1 \).

Here's how factorials work:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• For \( 0! \) and \( 1! \), both are specially defined to be 1.

Factorials grow very quickly as \( n \) increases, making larger computations easy to get complex. However, they form the backbone of combinatorial math, helping us in determinations of combinations and permutations.

Remember, a factorial is just a way of expressing multiplication over a range of numbers down to 1.

Recursive Algorithm: Solving by Going Smaller.

A recursive algorithm is like solving a mystery by breaking it down into smaller, more manageable mysteries. It's a method where the solution to a problem depends on solutions to smaller instances of the same problem.

Here's a look at recursive thinking in combinations:

• Start with \( C(n, r) = C(n-1, r-1) + C(n-1, r) \)

This equation suggests that to find how many ways to select \( r \) items from \( n \) items, you can:

• Select one, then choose \( r-1 \) from the remaining \( n-1 \).
• Or skip the selected item and choose \( r \) from the remaining \( n-1 \).

Recursion divides each problem into simpler sub-problems, eventually reaching a straightforward answer by using previously defined or smaller answers.

It’s a bit like peeling layers from an onion until you reach the core and solve the problem from there.

Base Case in Recursion: Your Stop Sign.

The base case in a recursive algorithm is what keeps it from running indefinitely and crashing. It's like a stop sign that tells your algorithm when to stop calculating deeper. Without it, any recursive algorithm would be like a car without brakes.

For combinations, the base case is defined as:

• \( C(n, 0) = 1 \)
• \( C(n, n) = 1 \)

These signify that:

• If no items are to be selected from \( n \) (that is \( r = 0 \)), there's only one way: select none.
• If all \( n \) items are to be selected, there's only one way: pick them all.

The base case ensures that our recursion doesn’t fall into an infinite loop, providing a condition from which it can start returning results upwards.