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Chapter 6: Problem 45

Using Theorem \(6.4,\) prove that \(P(n, r)=P(n-1, r)+r P(n-1, r-1)\) .

Short Answer

Expert verified
To prove that \(P(n, r)=P(n-1, r)+r P(n-1, r-1)\) using Theorem 6.4, we first recall that \(P(n, r) = nP(n-1, r-1)\) from the theorem. Then, we express \(P(n-1, r)\) using the theorem, finding \(P(n-1, r) = (n-1) P(n-2, r-1)\). We substitute these expressions into the left side of the equation and rewrite all terms as factorials, obtaining: \[\frac{n!}{(n-r)!} = \frac{(n-1)!}{(n-r)!} + r \frac{(n-1)!}{(n-r-1)!}\] This equation confirms the given result and demonstrates the relationship between \(P(n, r)\), \(P(n-1, r)\), and \(r P(n-1, r-1)\).

Step by step solution

01

Understanding Permutations

A permutation is an arrangement of objects from a set, where the order of the objects matters. In general, the number of permutations of n objects taken r at a time, denoted as \(P(n, r)\), can be calculated using the formula: \[ P(n, r) = \frac{n!}{(n-r)!} \] where '!' denotes the factorial, which is the product of all positive integers less than or equal to n.
02

State Theorem 6.4

Theorem 6.4 is a result that indicates the relationship between the number of permutations of n objects taken r at a time and the number of permutations of (n-1) objects taken r at a time. The theorem states: \[P(n, r) = n P(n-1, r-1)\]
03

Applying Theorem 6.4

We are asked to use Theorem 6.4 to prove that: \[P(n, r) = P(n-1, r) + r P(n-1, r-1)\] We know that \(P(n, r) = nP(n-1, r-1)\) from the theorem. To find the relationship between \(P(n, r)\) and \(P(n-1, r)\), we can derive \(P(n-1, r)\) by applying Theorem 6.4 for (n-1) objects: \[P(n-1, r) = (n-1)P(n-2, r-1)\] Now, we'll express \(P(n, r)\) in terms of \(P(n-1, r)\) and \(r P(n-1, r-1)\): \[nP(n-1, r-1) = P(n-1, r) + r P(n-1, r-1)\]
04

Solve for P(n, r)

Now we will solve for \(P(n, r)\), which can be written using the formula \(P(n, r) = \frac{n!}{(n-r)!}\). We can substitute \(nP(n-1, r-1)\) as follows: \[\frac{n!}{(n-r)!} = P(n-1, r) + r P(n-1, r-1)\] We can now rewrite \(P(n-1, r)\) and \(r P(n-1, r-1)\) in terms of factorials: \[\frac{n!}{(n-r)!} = \frac{(n-1)!}{(n-r)!} + r \frac{(n-1)!}{(n-r-1)!}\] At this point, we have established the relationship between \(P(n, r)\), \(P(n-1, r)\), and \(r P(n-1, r-1)\). The left side is equal to the right side in the equation above, and this confirms the given result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial
A factorial, denoted by the symbol '!', is a fundamental concept in permutations. The factorial of a number, say \( n! \), is the product of all positive integers less than or equal to \( n \). For example:
  • \( 3! = 3 \times 2 \times 1 = 6 \)
  • \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
Factorials are used to determine the number of ways to arrange a set of objects. In permutations, when you want to select \( r \) objects from \( n \) without replacement and where order matters, the formula is:\[ P(n, r) = \frac{n!}{(n-r)!} \]This formula calculates the number of permutations available by dividing the total arrangements \( n! \) by the arrangements of the remaining objects \((n-r)!\). This helps in solving many problems in discrete mathematics.
Theorem 6.4
Theorem 6.4 is a key result in permutations that gives us an essential relationship:\[ P(n, r) = n P(n-1, r-1) \]This formula tells us that the number of ways to arrange \( r \) objects from \( n \) is equal to the number of ways to select \( r-1 \) objects from \( n-1 \), multiplied by \( n \). The theorem is useful for breaking down permutation problems into smaller, manageable parts.
In proving expressions like \( P(n, r) = P(n-1, r) + r P(n-1, r-1) \), we apply Theorem 6.4 and transform these equations using known relations. By simplifying the permutations recursively, one can understand the layers of the problem, ultimately leading to a proof of more complex equations.
Combinatorial Proof
A combinatorial proof involves verifying an identity or proposition in mathematics using counting arguments. It's about demonstrating that both sides of an equation represent the same count through different perspectives or techniques.
For example, to prove \( P(n, r) = P(n-1, r) + r P(n-1, r-1) \), we can interpret both sides as counting different scenarios:
  • \( P(n-1, r) \) counts permutations where the first object is not used.
  • \( r P(n-1, r-1) \) counts permutations where the first object is used, multiplied by the remaining permutations.
These techniques often make abstract algebraic proofs more understandable by connecting them to tangible counting concepts.
Discrete Mathematics
Discrete mathematics is a branch of mathematics dealing with discrete elements that use algebra and arithmetic. It's foundational for computer science as it includes subjects like graph theory, logic, and set theory.
A core area of discrete mathematics is combinatorics, which includes the study of counting principles like permutations and combinations. By understanding permutations, for example, students learn to arrange objects where the order is significant, solving problems in scheduling, cryptography, and algorithms.
  • Permutations: Arranging objects where order matters.
  • Combinations: Selecting objects where order does not matter.
By mastering these principles, one gains the ability to tackle complex computational problems logically and systematically.

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