Problem 68 - Algebra 2

Question

a. Show that \((k+1) !=(k+1) \cdot k !\) b. Show that \(n \mathrm{C}_{k}+_{n} \mathrm{C}_{k}+1=n+1 \mathrm{C}_{k+1}\) c. Suppose \(n=4\) and \(k=2 .\) What entries in Pascal's Triangle are represented by \(_{n} \mathrm{C}_{k},_{n} \mathrm{C}_{k}+1,\) and \(_{n}+1 \mathrm{C}_{k}+1\) ? Verify that the equation in part (b) is true for these entries.

Step-by-Step Solution

Answer

a. The factorial equation is true by the definition of factorials. b. The combination equation is true by manipulating combination formula. c. The combination equation also holds true in the specific case of \(n = 4\) and \(k = 2\) in Pascal's Triangle.

Step 1 - Understanding Factorials

For part (a), recall that the factorial of \(k + 1\) or \((k + 1)!\) is defined as \((k + 1) \cdot k!\) where \(k!\) is the product of all positive integers less than or equal to \(k\). That is, \((k + 1)! = (k + 1) \cdot k!\), so part (a) is proven.

Step 2 - Manipulating Combiantions

For part (b), recall the definition of combinations: \(^n C_k = \frac{n!}{k!(n-k)!}\). So, \(^n C_k + ^n C_{k+1} = \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!}\) = \frac{n!(k+1+n-k)n! + k!n!}{k!(n-k)!(k+1)(n-k)} = \((n + 1)\frac{n!}{k+1!(n-k)!(n+1-k)} = ^{n+1} C_{k+1}\), which proves the equality given in part (b).

Step 3 - Pascal's Triangle

In Pascal's Triangle, each entry is the combination \(^n C_k\) where \(n\) is the row number (starting with 0) and \(k\) is the position (also starting with 0). For part (c) with \(n = 4\) and \(k = 2\), the entries are: \(^4 C_2 = 6, ^4 C_{2+1} = ^4 C_3 = 4\), and \(^{4+1} C_{2+1} = ^5 C_3 = 10\). The equality established in part (b), \(^n C_k + ^n C_{k+1} = ^{n+1} C_{k+1}\) or \(6 + 4 = 10\) holds true in this case.

Key Concepts

Factorial

The factorial is a fundamental concept in combinatorics, denoted by an exclamation mark (!). For any positive integer \( n \), the factorial \( n! \) is the product of all positive integers from 1 to \( n \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

It is essential to understand that the factorial of \( (k+1) \) can be expressed using the factorial of \( k \). This is because:

• \((k + 1)! = (k + 1) \times k!\)

This recursive property is incredibly handy, especially when simplifying expressions in problems involving permutations and combinations. Hence, part (a) of the exercise is effectively illustrating this recursive nature of factorials.

Pascal's Triangle

Pascal's Triangle is a stunning arithmetical tool with patterns that help solve combinatorial problems easily. It is structured like a pyramid, with each number representing a binomial coefficient. The topmost row begins with a 1, and each subsequent row has numbers that are the sum of the two directly above it.

For instance:

• Row 0: 1
• Row 1: 1 1
• Row 2: 1 2 1
• Row 3: 1 3 3 1

In part (c) of the exercise, \(n = 4\) and \(k = 2\), we want to locate the entries in Pascal's Triangle. The entries are \(^4C_2\), \(^4C_3\), and \(^5C_3\), represented by 6, 4, and 10, respectively. The triangle noticeably reveals the relationship and addition properties of combinations, confirming equations like those in part (b) of the exercise.

Binomial Coefficient

The binomial coefficient, denoted as \({_n}C_k\) or \(\binom{n}{k}\), represents the ways to choose \(k\) elements from \(n\) elements without regard to the order. The formula for binomial coefficients is:

• \({_n}C_k = \frac{n!}{k!(n-k)!}\)

This formula arises from counting the arrangements and dividing by the repeated arrangements for indistinguishable items.

In part (b) of the exercise, the equation \({_n}C_k + {_n}C_{k+1} = {_{n+1}}C_{k+1}\) is proven using the thoughts behind this formula. The exercise demonstrates that the sum of two adjacent binomial coefficients equals a binomial coefficient in the next row but one position to the right, reflecting the additive relationship visible in Pascal’s Triangle. Understanding this property helps in recognizing how combinatorics can provide solutions to complex problems smoothly.