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Chapter 29: Problem 2

Write out explicitly (a) \(\left(\begin{array}{l}n \\ 0\end{array}\right)\) (b) \(\left(\begin{array}{l}n \\ 1\end{array}\right)\) (c) \(\left(\begin{array}{l}n \\ 2\end{array}\right)\)

Short Answer

Expert verified
(a) 1, (b) n, (c) \( \frac{n(n-1)}{2} \)

Step by step solution

01

Understanding the Combination Notation

The notation \( \left( \begin{array}{c} n \ r \end{array} \right) \) represents the number of ways to choose \( r \) elements from a set of \( n \) elements. It is also known as a binomial coefficient and is calculated as \( \frac{n!}{r!(n-r)!} \), where \( ! \) denotes factorial.
02

Step A - Calculating \( \left( \begin{array}{c} n \\ 0 \end{array} \right) \)

For \( \left( \begin{array}{c} n \ 0 \end{array} \right) \), we calculate \( \frac{n!}{0!(n-0)!} \). Since \( 0! = 1 \), this simplifies to \( \frac{n!}{n!} = 1 \). Thus, \( \left( \begin{array}{c} n \ 0 \end{array} \right) = 1 \).
03

Step B - Calculating \( \left( \begin{array}{c} n \\ 1 \end{array} \right) \)

For \( \left( \begin{array}{c} n \ 1 \end{array} \right) \), we have \( \frac{n!}{1!(n-1)!} \). Since \( 1! = 1 \), this reduces to \( \frac{n!}{(n-1)!} = n \), which means \( \left( \begin{array}{c} n \ 1 \end{array} \right) = n \).
04

Step C - Calculating \( \left( \begin{array}{c} n \\ 2 \end{array} \right) \)

For \( \left( \begin{array}{c} n \ 2 \end{array} \right) \), the formula is \( \frac{n!}{2!(n-2)!} \). Simplifying, \( 2! = 2 \), this becomes: \( \frac{n \cdot (n - 1) \cdot (n-2)!}{2 \cdot (n-2)!} \). Canceling \( (n-2)! \), we get \( \frac{n(n-1)}{2} \). Thus, \( \left( \begin{array}{c} n \ 2 \end{array} \right) = \frac{n(n-1)}{2} \).

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Key Concepts

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

Understanding Binomial Coefficient
The binomial coefficient is a crucial concept in combinatorics. It helps us determine the number of ways to select a specific number of items from a larger set. This is denoted by the symbol \( \left( \begin{array}{c} n \ r \end{array} \right) \), where \( n \) is the total number of items, and \( r \) is the number of items to choose.
  • Meaning: The binomial coefficient tells us how many combinations of \( r \) items can be chosen from \( n \) without considering the order.
  • Calculation: It is calculated using the formula \( \frac{n!}{r!(n-r)!} \), which involves factorials of \( n \), \( r \), and \( n-r \).
  • Examples: Using this, we can easily compute \( \left( \begin{array}{c} n \ 0 \end{array} \right) = 1 \), meaning there's exactly one way to choose nothing (choose 0 items).
This formula provides the foundation for solving many problems in probability, statistics, and other mathematical fields.
The Functionality of Factorials
Factorials, denoted by an exclamation mark \( ! \), are essential for calculating permutations and combinations. The factorial of a number \( n \) is the product of all positive integers less than or equal to \( n \).
  • Basic Definition: For any positive integer \( n \), \( n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \).
  • Special Cases: By definition, we consider \( 0! = 1 \), which simplifies many calculations, particularly when \( r = 0 \) in binomial coefficients.
  • Usage in Problems: Factorials are used in the combination formula \( \frac{n!}{r!(n-r)!} \), simplifying expressions when finding combinations of elements.
Factorials grow rapidly, which can make calculations involving large numbers quite interesting!
Exploring the Combination Formula
The combination formula, built using binomial coefficients and factorials, is a powerful tool for solving problems where order does not matter in selections.
  • Formula: The formula is \( \left( \begin{array}{c} n \ r \end{array} \right) = \frac{n!}{r!(n-r)!} \). This expression tells us how to divide the total arrangements of \( n \) elements by the arrangements of \( r \) elements and the remaining \( (n-r) \) elements.
  • Applications: This formula is widely used in fields like probability, where determining the number of ways events can occur is crucial.
  • Step-by-Step Example: For \( \left( \begin{array}{c} n \ 1 \end{array} \right) = \frac{n!}{1!(n-1)!} \), simplifying gives \( n \), as it is simply choosing one item out of many.
These insights make complex problems more approachable and easier to solve systematically.

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Most popular questions from this chapter

A radar station has an \(M T B F\) of 1000 hours. The average repair time is 10 hours. Calculate the average availability of the radar station.

The mean of the numbers \(\left\\{x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right\\}\) is \(\bar{x}\). Show that the sum of the deviations about the mean is \(0 ;\) that is, show \(\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)=0\).

Two samples of nails are taken. The first sample has 12 nails with a mean length of \(2.7 \mathrm{~cm}\), the second sample has 20 nails with a mean length of \(2.61 \mathrm{~cm}\). What is the mean length of all 32 nails?

The probability distribution for the random variable, \(x\), is \begin{tabular}{lllllll} \hline\(x\) & 2 & \(2.5\) & \(3.0\) & \(3.5\) & \(4.0\) & \(4.5\) \\ \(P(x)\) & \(0.07\) & \(0.36\) & \(0.21\) & \(0.19\) & \(0.10\) & \(0.07\) \\ \hline (f) The variable, \(x\), is sampled 50000 times. How many times would you expect \(x\) to have a value of \(2.5 ?\) \end{tabular} (a) State \(P(x=3.5)\) (b) Calculate \(P(x \geqslant 3.0)\) (c) Calculate \(P(x<4.0)\) (d) Calculate \(P(x>3.5)\) (e) Calculate \(P(x \leqslant 3.9)\)

The probability a machine has a lifespan of more than 5 years is \(0.8\). Ten machines are chosen at random. What is the probability that (a) eight machines have a lifespan of more than 5 years (b) all machines have a lifespan of more than 5 years (c) at least eight machines have a lifespan of more than 5 years (d) no more than two machines have a lifespan of less than 5 years?
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