Problem 40 - Algebra 2

Question

Show that \(C(n-1, r)+C(n-1, r-1)=C(n, r)\)

Step-by-Step Solution

Answer

The equation is true, demonstrated by using Pascal's Rule.

Step 1: Understanding the Problem

We need to prove that the given equation \(C(n-1, r) + C(n-1, r-1) = C(n, r)\) holds true. Here, \(C(n, r)\) denotes a combination, which represents the number of ways to choose \(r\) elements from \(n\) elements.

Step 2: Defining Combinations

The combination formula is given by \(C(n, r) = \frac{n!}{r!(n-r)!}\), where \(n!\) is the factorial of \(n\). We'll use this formula to express each term in the equation.

Step 3: Applying the Combination Formula

First express each term using the combination formula: \(C(n-1, r) = \frac{(n-1)!}{r!(n-1-r)!}\) \(C(n-1, r-1) = \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!}\).

Step 4: Simplifying Factorials

Simplify the factorial expressions: \[(n-1-r)! = (n-r-1)!,\]\[(n-1-(r-1))! = (n-r)!,\]Thus, the expressions become: \[C(n-1, r) = \frac{(n-1)!}{r!(n-r-1)!},\]\[C(n-1, r-1) = \frac{(n-1)!}{(r-1)!(n-r)!}.\]

Step 5: Adding the Combinations

Add the expressions \(C(n-1, r)\) and \(C(n-1, r-1)\):\[C(n-1, r) + C(n-1, r-1) = \frac{(n-1)!}{r!(n-r-1)!} + \frac{(n-1)!}{(r-1)!(n-r)!}.\]

Step 6: Aligning Common Denominator

To add these fractions, find a common denominator, which is \(r!(n-r)!\):\[\frac{(n-1)!}{r!(n-r-1)!} = \frac{(n-1)! imes (n-r)}{r!(n-r)!},\]\[\frac{(n-1)!}{(r-1)!(n-r)!} = \frac{(n-1)! imes r}{r!(n-r)!}.\]

Step 7: Combining Fractions

Now, add the two fractions with a common denominator:\[\frac{(n-1)! imes (n-r) + (n-1)! imes r}{r!(n-r)!} = \frac{(n-1)!\cdot(n-r + r)}{r!(n-r)!}.\]

Step 8: Simplifying the Sum

The numerator simplifies to:\((n-1)!\times (n) = n!\)Thus forming:\[\frac{n!}{r!(n-r)!},\] which is precisely \(C(n, r)\).

Step 9: Conclusion

Therefore, we have shown that \[C(n-1, r) + C(n-1, r-1) = C(n, r).\]This is known as Pascal's Rule for combinations.

Key Concepts

Combinatorics

Combinatorics is a fascinating branch of mathematics that deals with counting, arranging, and grouping objects. This field helps us determine the number of possible outcomes in various scenarios. For instance, when we want to know how many ways we can form a committee from a group, or rearrange letters in a word, we use combinatorial methods. Combinatorics encompasses a variety of topics, such as permutations, combinations, and the principle of inclusion-exclusion. Each of these topics allows us to solve different types of counting problems involving sets and sequences.

• Permutations focus on arranging objects where order matters.


• Combinations focus on selecting objects where order does not matter.


• The principle of inclusion-exclusion helps when groups overlap.

In this exercise, we focus on combinations, which are used to understand how many subsets or groups can be formed from a larger set without regard to the order of elements.

Binomial Coefficients

Binomial coefficients are crucial in combinatorics for counting the number of ways to select a subset of items from a larger set. They are often written as \(C(n, r)\) or \(\binom{n}{r}\), and are pronounced as "n choose r". This represents the number of combinations possible when choosing \(r\) items from \(n\) items. The formula for binomial coefficients is \(C(n, r) = \frac{n!}{r!(n-r)!}\).

• "\(n!\)" (n factorial) represents the total ways to arrange \(n\) items.


• "\(r!\)" accounts for different arrangements within the chosen group.


• "\((n-r)!\)" eliminates the counts of remaining unchosen items.

Binomial coefficients have many applications, such as computing probabilities, and are fundamental in binomial expansions. For example, in Pascal's Rule, they help establish relationships between different coefficients, providing a simplified way to calculate combinations recursively.

Factorials

Factorials are a building block of combinatorics and are essential in calculations involving permutations and combinations. The factorial of a positive integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

• Factorials count the number of ways to arrange \(n\) distinct objects.


• The factorial function grows very quickly; even for small \(n\), values can become quite large.


• By convention, \(0! = 1\), as there is one way to arrange zero items: do nothing.

Factorials are integral to understanding combinations and permutations. In the context of the binomial coefficients, they help simplify the counting of arrangements and allow for the derivation of recursive properties like those found in Pascal's Rule.