Problem 40 - Discrete Mathematics with Applications

Question

Using the recursive definition of \(P(n, r),\) evaluate each. $$P(6,3)$$

Step-by-Step Solution

Answer

Using the recursive definition of \(P(n, r)\), we evaluate \(P(6, 3)\) as follows: \(P(6, 3) = 6 \cdot P(5, 2) = 6 \cdot (5 \cdot P(4, 1)) = 6 \cdot (5 \cdot 4) = 6 \cdot 20 = 120\). Therefore, \(P(6, 3) = 120\).

Step 1: Apply the Recursive Formula

Apply the recursive formula for permutations to calculate \(P(6, 3)\). Here, \(n = 6\) and \(r = 3\). Therefore, we have: \(P(6, 3) = 6 \cdot P(6-1, 3-1)\)

Step 2: Simplify the Expression

Simplify the expression inside the parentheses: \(P(6, 3) = 6 \cdot P(5, 2)\)

Step 3: Apply the Recursive Formula again

Apply the recursive formula for permutations again to calculate \(P(5, 2)\). This time, we have \(n = 5\) and \(r = 2\): \(P(5, 2) = 5 \cdot P(5-1, 2-1)\)

Step 4: Simplify the Expression

Simplify the expression inside the parentheses: \(P(5, 2) = 5 \cdot P(4, 1)\)

Step 5: Simplify the Base Case

We have reached the base case, where \(r = 1\). When \(r = 1\), we can simplify the expression to \(n\): \(P(4, 1) = 4\)

Step 6: Substitute and Multiply

Now that we have simplified the base case, we can substitute it back into the previous steps and multiply: \(P(5, 2) = 5 \cdot 4\) \(P(5, 2) = 20\) \(P(6, 3) = 6 \cdot 20\) \(P(6, 3) = 120\)

Step 7: State the Final Answer

Now we have calculated \(P(6, 3)\) using the recursive definition: \(P(6, 3) = 120\)

Key Concepts

Recursive Definition

recursively until ( (n, 1) = n, which is your base case. In other words, if you are looking for the permutation of n items taken one at a time, it's simply n, since there's only one way to arrange one item.

Factorial Notation

Factorial notation is a mathematical concept used to describe the product of an integer and all the non-zero integers below it. It is denoted by an exclamation mark (!). For example, 4! (read as 'four factorial') is calculated as 4 x 3 x 2 x 1. Factorials are integral in combinatorial analysis, particularly in permutations and combinations.

When the term permutation comes into play, we often use factorial to express the total number of distinct sequences. If we want to consider all possible orders for n items, the notation becomes n!, which incorporates all permutations of the set. This factorial concept paves the way for understanding the permutations formula better, as it is closely tied with the recursive definition of permutations.

Combinatorial Analysis

Combinatorial analysis is the field of mathematics focused on counting, evaluating, and predicting the possible outcomes of various arrangements of collections of objects. It encompasses combinations and permutations as two of its most crucial concepts.

In permutations, the order of arrangement matters, which makes them distinct from combinations where the order is irrelevant. Combinatorial analysis not only addresses the number of possible outcomes but also aids in strategic decision-making, probability theory, and optimization. Understanding the principles and formulas of combinatorial analysis is valuable for students in a myriad of academic and professional fields.

Permutations Formula

The permutations formula is used to determine the number of ways r items can be ordered from a set of n unique items. The classic formula is ( (n, r) = n! / (n-r)!, where n is the total number of items, and r is the number of items to arrange.

The formula tells us how many different ways we can pick and order r items out of n available. In our example of finding the number of ways to give out the top three places in a race with six runners, we use the permutations formula as follows: ( P(6, 3) = frac{6!}{(6-3)!}, which calculates to 120 distinct ways. The permutations formula is a powerful tool because it generalizes the problem of ordering items and provides a method to systematically calculate permutations, bypassing the need for enumeration of each configuration.