Problem 81 - A Graphical Approach to College Algebra

Question

Prove each statement for positive integers \(n\) and \(r,\) with \(r \leq n\) (Hint: Use the definitions of permutations and combinations.) $$C(n, n-1)=n$$

Step-by-Step Solution

Answer

The statement is true since \( C(n, n-1) = n \) by definition of combinations.

Step 1: Understand Combinations

The combination formula is given by \( C(n, r) = \frac{n!}{r!(n-r)!} \), which represents the number of ways to choose \( r \) elements from a set of \( n \) elements without regard to order.

Step 2: Plug Values into Combination Formula

To prove \( C(n, n-1) = n \), we substitute \( r = n-1 \) into the combination formula: \[ C(n, n-1) = \frac{n!}{(n-1)!(n-(n-1))!} = \frac{n!}{(n-1)! \cdot 1!} \]

Step 3: Simplify the Expression

Simplify \( \frac{n!}{(n-1)! \cdot 1!} \): \[ \frac{n!}{(n-1)! \cdot 1!} = \frac{n \cdot (n-1)!}{(n-1)!} \cdot 1 = n \] By cancelling \((n-1)!\), we are left with \( n \), which verifies the statement.

Key Concepts

Permutations

Permutations are a fundamental concept in combinatorics, where the order of arrangement is taken into consideration. Imagine you're arranging books on a shelf, and the sequence matters. Each unique sequence is a permutation. Mathematically, a permutation of a set of elements is the number of ways to arrange these elements in a specific order.

The formula to determine the number of permutations of items taken at a time is:
\[P(n, r) = \frac{n!}{(n-r)!}\]
This formula is derived from the idea that you have choices for the first position, choices for the second, and so on, until positions are filled.

Permutations differ from combinations, as they concern the arrangement order. For instance, arranging 3 distinct letters A, B, C results in six permutations: ABC, ACB, BAC, BCA, CAB, CBA.

• Order matters in permutations.
• Use the permutations formula when sequence is important.

Combinations

Combinations focus on the selection of items from a larger pool, where the order does not matter. This is akin to selecting a committee from a larger group, where it doesn't matter in which order the committee members were chosen.

The formula for calculating the combinations of items taken at a time is:
\[C(n, r) = \frac{n!}{r!(n-r)!}\]
This formula considers that each selection of items could be arranged in r factorial (!) different ways, which is why we divide by ! in the formula to avoid overcounting orders.

The problem given, attempts to prove that selecting items out of , results in possible combinations. By using the combination formula, substituting for and for -1, the solution verifies that selecting all but one item from a set results in combinations - a straightforward but insightful result.

• Order does not matter in combinations.
• The formula accounts for all possible arrangements, hence the division by factorial.

Factorials

Factorials are mathematical functions often used in permutations and combinations. A factorial, indicated by the symbol !, represents the product of all positive integers up to that number. For example, 5! is calculated as follows:
\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]

Factorials are vital in calculating permutations and combinations, as they help manage all possible arrangements of items which are either considered or omitted based on context. In permutations, it quantifies the possible orderings of from , while in combinations, it ensures the calculation reflects the unordered nature of the group selection.

Factorials also have special considerations:

• By convention, 0! equals 1, because there’s only one way to arrange zero items: do nothing.


• They grow very quickly; hence often involved in simplifications like cancelling out in division during calculations.