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Chapter 10: Problem 17

Evaluate \(C(n, r)\). $$ C(4,2) $$

Short Answer

Expert verified
There are 6 ways to choose 2 items from a set of 4.

Step by step solution

01

Understand Combination

The expression \( C(n, r) \) represents the number of combinations, or ways, to choose \( r \) items from \( n \) items without regard to order. It is calculated using the formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
02

Identify Variables

In this problem, we have \( n = 4 \) and \( r = 2 \). These values represent choosing 2 items from a set of 4.
03

Calculate Factorials

Calculate the factorial values: \( 4! = 4 \times 3 \times 2 \times 1 = 24 \), \( 2! = 2 \times 1 = 2 \), and \( (4-2)! = 2! = 2 \).
04

Substitute Values

Substitute the factorial values into the combination formula: \( C(4, 2) = \frac{4!}{2! (4-2)!} = \frac{24}{2 \times 2} \).
05

Simplify

Simplify the expression: \( \frac{24}{4} = 6 \). This means there are 6 ways to choose 2 items from 4.

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Key Concepts

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

Understanding Factorials
When we talk about factorials, we're dealing with a special type of multiplication sequence that plays a vital role in counting methods like combinations and permutations. A factorial, denoted by an exclamation mark (!), is the product of an integer and all the integers below it down to one. For example:
  • 4! means multiplying all whole numbers from 4 down to 1: \[4! = 4 \times 3 \times 2 \times 1 = 24\]
  • Similarly, 2! ("2 factorial") would be: \[2! = 2 \times 1 = 2\]
  • By convention, 0! is defined to be 1.
Bringing this into combinations, the concept of factorial allows us to determine the number of ways items can be arranged or chosen. In combinations, where order doesn鈥檛 matter, factorials help manage how we divide arrangements into distinct sets.
Decoding the nCr Formula
The notation \(C(n, r)\) represents combinations, which is used to pick \( r \) items from \( n \) total items, disregarding the sequence of selection. The formula to calculate this is:
\[C(n, r) = \frac{n!}{r!(n-r)!}\]
Here's how it breaks down:
  • **Numerator**: \(n!\) represents the number of ways to arrange \(n\) items.
  • **Denominator**: \(r!\times(n-r)!\) adjusts for overcounting since the order doesn鈥檛 matter.
So, to find \(C(4, 2)\), we'd compute:
- \(4!\) for all arrangements of 4 items.- \(2!\) for the selected items, and \((4-2)!\) for the unselected.
When you substitute and simplify:
\[C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{24}{4} = 6\]Therefore, there are 6 possible combinations or ways to choose 2 items out of 4.
Permutations Versus Combinations
Understanding the difference between permutations and combinations is crucial when dealing with arrangements of items. Both concepts involve selecting items from a larger set but do so with a key difference:
  • **Permutations** account for order. The position of each item matters, which means that rearrangements of the same items are counted separately. For example, choosing and then arranging 2 books out of a set of 4 is a matter of permutations.
  • **Combinations**, however, do not account for order. Here, only the selection matters, not the sequence. Choosing 2 ice cream flavors out of 4 options leaves the order of selection unimportant.
For example, the permutations formula is \(P(n, r) = \frac{n!}{(n-r)!}\) while the combinations formula is \(C(n, r) = \frac{n!}{r!(n-r)!}\). Notice how combinations divide by additional \(r!\) to disregard order, making it a "safer" choice for problems where arrangement doesn鈥檛 matter.

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