Problem 2 - Algebra 2 and Trigonometry

Question

Hunter said that the number of combinations of \(n\) things taken \(r\) at a time is equal to the number of permutations of \(n\) things taken \(n\) at a time when \(r\) are identical; that is, \(C_{r}=\frac{n !}{r !}\) . Do you agree with Hunter? Explain why or why not.

Step-by-Step Solution

Answer

Hunter is incorrect because he confused permutations with combinations; his formula would only apply if all objects were identical or when \( n = r \).

Step 1: Understanding Combinations

The number of combinations of \( n \) things taken \( r \) at a time is given by the formula \( \binom{n}{r} \). This is expressed as \( \frac{n!}{r!(n-r)!} \). Here, \( n! \) is the factorial of \( n \), which calculates the product of all positive integers up to \( n \). The formula for combinations accounts for selections where the order does not matter.

Step 2: Understanding Permutations with Identical Objects

Permutations involve arrangements of objects where order matters. When \( r \) objects are identical among \( n \) total objects, the formula used is \( \frac{n!}{r!} \). This formula is used because the identical objects don't contribute to different orders, so we divide by \( r! \) to account for these repetitions.

Step 3: Analysis of Hunter's Statement

Hunter states that \( C_{r} = \frac{n!}{r!} \). Given \( C_{r} \) implies the combination formula \( \binom{n}{r} \), which should be \( \frac{n!}{r!(n-r)!} \), Hunter's statement fails to account for the \( (n-r)! \) in the denominator. The correct formula for combinations does not match \( \frac{n!}{r!} \) unless \( n = r \), which reduces the combination formula to \( 1 \).

Step 4: Conclusion

Hunter's statement is incorrect for calculating combinations with varying selections because it only applies when permutation calculations are considered with identical objects. The statement equivalently discusses permutations, not combinations, as usually expressed.

Key Concepts

Permutations

Permutations are all about arranging objects where the order is important. Imagine playing with building blocks. If you rearrange them, you get a new order each time. That’s what permutations deal with.
In mathematical terms, if you have a collection of objects and want to know how many ways you can arrange them, you use permutations. The formula for finding the number of permutations of a set of values is given by:

• For arranging all objects: \(n!\)
• For arranging a subset: \(\frac{n!}{(n-r)!}\)

The '! ' symbol, known as a factorial, represents multiplying a series of descending natural numbers. For example, \(5!\) is \(5 \times 4 \times 3 \times 2 \times 1\).
It’s important to note that permutations change as items are arranged, and different orders are essential. Every new arrangement counts as a new permutation.

Factorials

Factorials might seem complex, but they’re simpler than you think. Factorials are what you get when you multiply a series of descending numbers. It’s kind of like counting but in multiplication terms.
For instance:

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

The concept of factorials is essential in permutations and combinations because it helps calculate how many different ways things can be arranged or chosen. It provides a quick and efficient method to solve complex arrangements without counting each possibility manually.
Beyond permutations, factorials appear in various mathematical situations, indicating their versatility. They are a critical tool in the mathematician's toolkit for arranging and selecting items.

Identical Objects

Identical objects pose a special case when arranging or selecting items. When some items are identical, it doesn’t matter if you swap these identical items, the overall arrangement remains unchanged.
This impacts calculations in permutations. With identical objects, say \(r\) identical objects among \(n\) total objects, the formula for permutations becomes:

• \(\frac{n!}{r!}\)

By dividing with \(r!\), we effectively eliminate repeating sequences since swapping identical items doesn’t yield a new sequence. Consider a scenario: arranging the letters in the word "BALLOON." If 'L's or 'O’s are swapped, the arrangement visually and physically remains the same.
Understanding permutations with identical objects helps us avoid overcounting and ensures precise calculations of unique arrangements. So, this plays a key role in accurately solving problems involving repeat elements in permutations.