Problem 69 - Contemporary Precalculus

Question

Let \(r\) and \(n\) be integers such that \(0 \leq r \leq n\) (a) Verify that \((n-r) !=(n-r)[n-(r+1)] !\) (b) Verify that \((n-r) !=[(n+1)-(r+1)] !\) (c) Prove that \(\left(\begin{array}{c}n \\\ r+1\end{array}\right)+\left(\begin{array}{c}n \\\ r\end{array}\right)=\left(\begin{array}{c}n+1 \\ r+1\end{array}\right)\) for any \(r \leq n-1\) [Hint: Write out the terms on the left side and use parts (a) and (b) to express each of them as a fraction with denominator \((r+1) !(n-r) !\). Then add these two fractions, simplify the numerator, and compare the result with \(\left.\left(\begin{array}{l}n+1 \\ r+1\end{array}\right) .\right]\) (d) Use part (c) to explain why each entry in Pascal's triangle (except the l's at the beginning or end of a row) is the sum of the two closest entries in the row above it.

Step-by-Step Solution

Answer

Question: Verify and prove various properties of factorials and combinations, and connect the findings with Pascal's Triangle. Answer: We have verified the given properties of factorials and proven that \(\left(\begin{array}{c}n \\\ r+1\end{array}\right)+\left(\begin{array}{c}n \\\ r\end{array}\right)=\left(\begin{array}{c}n+1 \\\ r+1\end{array}\right)\) for any \(r \leq n-1\). This relationship between combinations illustrates the connection between factorials, combinations, and Pascal's Triangle, as each entry in a row of Pascal's Triangle is the sum of the two closest entries in the row above it (except for the 1's at the beginning or end of a row). This showcases the deep connection between these mathematical concepts and helps explain the patterns seen in Pascal's Triangle.

Step 1: Isolate n-r term

We can rewrite \((n-r) !\) as \((n-r)[n-(r+1)] !\) by isolating the term \((n-r)\).

Step 2: Verify the equation

We can see that the given equation \((n-r) !=(n-r)[n-(r+1)] !\) is indeed true, as we have just rewritten \((n-r) !\) in another form with the same value. (b) Verify that \((n-r) !=[(n+1)-(r+1)] !\)

Step 1: Simplify the equation

First, simplify the right side of the equation: \([(n+1)-(r+1)] = n+1-r-1 = (n-r)\).

Step 2: Verify the equation

Since the right side simplifies to \((n-r)\), the equation \((n-r) != (n-r)!\) is true. (c) Prove that \(\left(\begin{array}{c}n \\\ r+1\end{array}\right)+\left(\begin{array}{c}n \\\ r\end{array}\right)=\left(\begin{array}{c}n+1 \\\ r+1\end{array}\right)\) for any \(r \leq n-1\).

Step 1: Write out the terms

Using the definition of combinations, rewrite the terms as: \(\frac{n!}{(r+1)!(n-r)!} + \frac{n!}{r!(n-r)!}\).

Step 2: Find a common denominator

Find a common denominator for the two fractions: \((r+1)!(n-r)!\). Rewrite the fractions with this common denominator: \(\frac{n!(r+1)}{(r+1)!(n-r)!} + \frac{n!r}{r!(n-r)!}\).

Step 3: Simplify and compare

Add the two fractions: \(\frac{n!(r+1) + n!r}{(r+1)!(n-r)!}\). Factor out \(n!\): \(\frac{n![(r+1) + r]}{(r+1)!(n-r)!}\). This simplifies to \(\frac{(n+1)!}{(r+1)!(n-r)!}\), which is equivalent to \(\left(\begin{array}{c}n+1 \\\ r+1\end{array}\right)\). Thus, the proof is complete. (d) Use part (c) to explain why each entry in Pascal's triangle (except the 1's at the beginning or end of a row) is the sum of the two closest entries in the row above it.

Step 1: Relate the result to Pascal's Triangle

Each entry in Pascal's Triangle is a combination of the form \(\left(\begin{array}{c}n\\\ r\end{array}\right)\), where \(n\) is the row number (starting from 0), and \(r\) is the position of the entry within that row (also starting from 0).

Step 2: Connect the proof in part(c) with Pascal's Triangle

The result in part (c) shows that each entry in a row of Pascal's Triangle (except the 1's at the beginning or end of a row) is the sum of the two closest entries in the row above it. This is because \(\left(\begin{array}{c}n \\\ r+1\end{array}\right)+\left(\begin{array}{c}n \\\ r\end{array}\right)=\left(\begin{array}{c}n+1 \\\ r+1\end{array}\right)\), which is the relationship between entries in adjacent rows of Pascal's Triangle.

Key Concepts

Factorial Notation in Precalculus

Factorial notation is a fundamental concept in mathematics, especially in the field of combinatorics and precalculus. It is written as a positive integer followed by an exclamation point, such as 5!, and is defined as the product of all positive integers up to that number. For instance, the factorial of 5, denoted as 5!, is calculated by multiplying all positive integers from 1 to 5:
\begin{align*}5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\text{.}\begin{align*}When the factorial operation is used in combinatorics, it helps us count the number of ways to arrange or select items. In the exercise at hand, understanding factorial notation is crucial when interpreting combinations, where we find formulas such as \(\frac{n!}{r!(n-r)!}\) in Pascal's Triangle. It is also important in simplifying expressions and establishing relationships between factorial terms, as evidenced by the steps in verifying equations like \((n-r)! = (n-r)[n-(r+1)]!\).

Understanding Combinatorics

Combinatorics is the area of mathematics that focuses on counting, arrangement, and combination of things. It lies at the heart of problems involving probability, statistics, and discrete mathematics. One of the key elements of combinatorics is understanding how to count without actually having to list every possibility, which can be done using methods such as the rule of product, permutations, and combinations.
In the given exercise, when we talk about proving equations that involve combinations such as \(\left(\begin{array}{c}n \ r\end{array}\right)\), we're referring to the number of ways to choose r items out of n possible items without regard to order. The solution to exercise part (c) demonstrates how understanding the principles in combinatorics helps in recognizing patterns like those found in Pascal's Triangle. Identifying these patterns allows for the deep understanding of binomial expansion and probability calculations, key areas within precalculus.

Combinations and Permutations

While both combinations and permutations deal with the selection of items from a set, they differ in how they consider the order of selection. A permutation is an arrangement of items in a specific ordered sequence, whereas a combination is a selection of items where the order is not important.
The fundamental principle for calculating permutations is to count the number of ways to arrange r items from a set of n, which is given by the formula \(P(n, r) = \frac{n!}{(n-r)!}\). On the other hand, combinations are calculated using \(C(n, r) = \frac{n!}{r!(n-r)!}\), which is also represented by \(\left(\begin{array}{c}n \ r\end{array}\right)\) and embodies the entries of Pascal's Triangle.
The exercise we're examining showcases the relevance of combinations. When proving the relationship in Pascal's Triangle, we see that each number is a combination, reflecting the number of ways to select subsets of items from a set and further implying why, in Pascal's Triangle, we add the values above to get the new value below. This also emphasizes the symmetry within the triangle, where \(\left(\begin{array}{c}n \ r\end{array}\right)\) is equal to \(\left(\begin{array}{c}n \ n-r\end{array}\right)\), due to the nature of combinations disregarding order.