Problem 22 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the given expression. $$ C(n, n-2) $$

Step-by-Step Solution

Answer

The evaluated expression for \(C(n, n-2)\) is \(\frac{n (n-1)}{2}\).

Step 1: Substitute the value of r in the formula

Replace \(r\) with \(n-2\) in the combination formula: $$ C(n, n-2) = \frac{n!}{(n-2)!(n-(n-2))!} $$

Step 2: Simplify the expression

Now, simplify the expression by performing the operations inside the parenthesis: $$ C(n, n-2) = \frac{n!}{(n-2)!(2)!} $$

Step 3: Write the expression in terms of factorials

Recall that \(n! = n \times (n-1) \times (n-2)! \) and \(2! = 2 \times 1 = 2\). Now, rewrite the expression in terms of factorials: $$ C(n, n-2) = \frac{n \times (n-1) \times (n-2)!}{(n-2)! \times 2} $$

Step 4: Simplify the expression

Cancel out the common terms \((n-2)!\) in the numerator and denominator, which leaves us with: $$ C(n, n-2) = \frac{n \times (n-1)}{2} $$ Now, the expression has been evaluated, and we have found that: $$ C(n, n-2) = \frac{n (n-1)}{2} $$

Key Concepts

Combination Formula

When dealing with combinatorics, particularly in problems where selection matters, the combination formula serves as a crucial tool. It helps us determine how many ways we can choose a group of items from a larger set, without considering the order of selection. The general form of the combination formula is:

• \(C(n, r) = \frac{n!}{r!(n-r)!}\)

Here, \(n\) represents the total number of items, and \(r\) is the number of items to choose. Combining items without regard to order expresses itself perfectly through this formula. For instance, choosing a subcommittee from a larger committee without worrying who gets selected first is a typical scenario.The beauty of combinations lies in their simplicity and wide range of applications in probability, statistics, and everyday problems where grouping matters but order does not.

Factorials

Factorials play a pivotal role in the world of combinatorics, providing a way to simplify calculations involving permutations and combinations. Represented by the symbol \(!\), a factorial takes a non-negative integer \(n\) and computes the product of all whole numbers from \(n\) down to 1:

• \(n! = n \times (n-1) \times (n-2) \times \ldots \times 1\)

Factorials are fundamental for calculating combinations and permutations because they account for the various orders and groupings of a set.For example:

• \(3! = 3 \times 2 \times 1 = 6\)
• \(4! = 4 \times 3 \times 2 \times 1 = 24\)

The concept extends to 0! as well, where 0! is defined to be 1, allowing formulas for combinations and permutations to remain consistent.Understanding factorials is essential because they compress complex multiplicative combinations into manageable terms, facilitating more accessible problem-solving experience.

Binomial Coefficients

Among the building blocks of combinatorics are binomial coefficients, often appearing in problems dealing with the selection of groups. Found within Pascal's Triangle and represented with the "choose" notation \(C(n, r)\), they denote the number of ways to pick \(r\) items from \(n\) items without regard to order:

• \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)

This term, also referred to as "n choose r," finds itself naturally embedded in the binomial theorem, which expands algebraic expressions raised to a power. In essence, binomial coefficients provide the links between algebra, number theory, and combinatorics.For example, in expanding

• \((x + y)^2 = x^2 + 2xy + y^2\)

The coefficient 2 comes from the binomial coefficient \(\binom{2}{1}\), teaching us how multiple combinations of \(x\) and \(y\) align.In everyday decision-making, like picking committees or forming teams out of larger groups, binomial coefficients simplify the understanding of possible selections, acting as a guide through the complex world of possibilities.