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Magazine Chapter 9: Problem 37
Can \(C(n, r)\) ever equal \(P(n, r) ?\) Explain.
Short Answer
Expert verified
Yes, when \(r = 0\) or \(r = 1\).
Step by step solution
01
Understanding Combinations and Permutations
Let's first review the definitions. The combination formula, denoted as \(C(n, r)\), represents the number of ways to choose \(r\) items from \(n\) items without regard to the order and is calculated as \[C(n, r) = \frac{n!}{r!(n-r)!}\]. On the other hand, the permutation formula, denoted as \(P(n, r)\), represents the number of ways to choose \(r\) items from \(n\) items with regard to the order and is calculated as \[P(n, r) = \frac{n!}{(n-r)!}\].
02
Setting the Equations Equal
To find if the two can be equal, set the expressions for \(C(n, r)\) and \(P(n, r)\) equal to each other: \[\frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!}\]. This equation implies that the factorial terms must simplify to allow equality.
03
Simplifying the Equation
Cancel \(n!\) and \((n-r)!\) from both sides, leaving \[\frac{1}{r!} = 1\]. This simplification leads us to examine the factorial of \(r\).
04
Examining Factorial Condition
The equation \(\frac{1}{r!} = 1\) holds true when \(r! = 1\). The factorial of a number equals 1 only when \(r = 0\) or \(r = 1\), since \(0! = 1! = 1\). This means that the two expressions can only be equal when \(r\) is equal to these values.
05
Conclusion
Thus, \(C(n, r) = P(n, r)\) only when \(r = 0\) or \(r = 1\). In these scenarios, the position or ordering of the objects does not matter as there is either nothing to arrange or only one item to place.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Combinations
Combinations are all about selecting items without paying attention to the order. When you're looking at combinations, it's like picking a team. The lineup doesn't matter鈥攐nly who's in it. Mathematically, the combination of choosing \(r\) items from \(n\) items is given by the formula:
- The formula is \(C(n, r) = \frac{n!}{r!(n-r)!}\)
- Here, \(!\) means "factorial," which we will explain further later.
Permutations
Permutations focus on the arrangement of items, where the order suddenly becomes the star. Imagine you're arranging books on a shelf or setting up chairs around a table. Here, the sequence matters. For permutations, the formula is:
- \(P(n, r) = \frac{n!}{(n-r)!}\)
Factorial
Factorials are a building block in combinatorics. The factorial of a number, noted as \(!\), means multiplying it by every whole number below it down to 1. It's like a reverse countdown:
- \(n! = n \times (n-1) \times (n-2) \times \ldots \times 1\)
- Special cases are \(0! = 1! = 1\)
Order
Order is what separates permutations from combinations. With order, it's not just who or what you have, but how they line up or are placed. Order matters for:
- Sequences or seating arrangements
- Arrangement of digits in a number
Arrangement
Arrangement is about how items are lined up or organized and is closely tied to the concept of order. Think about how diverse arrangements can be pivotal:
- In meetings: who speaks when
- In sports: positioning of players
