/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 In how many ways can we choose f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 2

In how many ways can we choose five people from a group of ten to form a committee?

Short Answer

Expert verified
There are 252 ways to choose 5 people from a group of 10.

Step by step solution

01

Understanding the Problem

We need to find out how many different ways we can choose 5 people from a group of 10 people. This is a typical combination problem, where the order of choosing does not matter.
02

Applying Combination Formula

Use the combination formula to calculate the number of ways to choose 5 people from 10. The formula for combinations is \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
03

Calculating Factorials

Calculate the factorial for each part of the formula: \( 10! \), \( 5! \), and \( (10-5)! = 5! \). Recall that factorial means the product of all positive integers up to that number.
04

Substituting into the Formula

Substitute the values into the combination formula: \( C(10, 5) = \frac{10!}{5!5!} \). Now compute \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \) and notice that the \( 5! \) terms will cancel out in numerator and denominator.
05

Simplifying the Expression

After cancelling \( 5! \) from the numerator and denominator, you'll have \( C(10, 5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \). Simplify this to calculate the final number.
06

Final Calculation

Compute the multiplication: \( 10 \times 9 \times 8 \times 7 \times 6 = 30240 \) in the numerator and \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \) in the denominator. Then divide \( 30240 \div 120 = 252 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Factorials
Factorials are a mathematical concept used widely in problems involving permutations and combinations.
The factorial of a non-negative integer, denoted as \( n! \) (read as "n factorial"), is the product of all positive integers less than or equal to \( n \). Here's a simple breakdown to make it more understandable:
  • \( 0! = 1 \) by definition.
  • \( 1! = 1 \).
  • \( 3! = 3 \times 2 \times 1 = 6 \).
  • \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factorials grow very quickly. For example, \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800 \). This escalating nature makes them useful for counting different possible arrangements or sequences of items.
Exploring the Combination Formula
The combination formula is essential when solving problems that require selecting items or people without regard to order. It's mathematically expressed as:
\[C(n, r) = \frac{n!}{r!(n-r)!}\]This tells us how many ways we can choose \( r \) items from a total of \( n \) items, where the order doesn鈥檛 matter. Let's unfold this a bit for a better understanding.
  • \( n \) represents the total number of items. For instance, in our original problem, \( n = 10 \) signifies choosing from 10 people.
  • \( r \) represents the number of chosen items. Our problem specifies \( r = 5 \) since we want to choose 5 people for a committee.
  • Plugging values into the equation, we have \( C(10, 5) = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} \).
The formula does the hard work, highlighting the mechanism of choosing without concern for sequence, aligning perfectly with committee selection and similar scenarios.
Permutations vs. Combinations
Knowing the difference between permutations and combinations is crucial as both involve arrangements or selections, yet they handle order differently.
Permutations relate to the arrangement of items where order is key. When the sequence matters鈥攕ay, arranging books on a shelf鈥攑ermutations are applied. The permutation formula is given by:
\[P(n, r) = \frac{n!}{(n-r)!}\]In contrast, combinations disregard order, focusing purely on selection. Imagine forming teams or committees from a larger group, just like our original exercise. In such cases, two different orders of the same group (e.g., \{A, B, C\} and \{C, B, A\}) are considered the same combination.
To summarize:
  • Use permutations when the arrangement is important.
  • Use combinations when the grouping or selection order does not matter, as demonstrated with the committee formation exercise.
Recognizing when to apply these concepts is vital for tackling a wide array of mathematical and real-world problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Given any ordering \(\sigma\) of \(\\{1,2, \ldots, n\\},\) we can define \(\sigma^{-1},\) the inverse ordering of \(\sigma,\) to be the ordering in which the \(i\) th element is the position occupied by \(i\) in \(\sigma\). For example, if \(\sigma=(1,3,5,2,4,7,6)\), then \(\sigma^{-1}=(1,4,2,5,3,7,6)\). (If one thinks of these orderings as permutations, then \(\sigma^{-1}\) is the inverse of \(\left.\sigma .\right)\) A fall occurs between two positions in an ordering if the left position is occupied by a larger number than the right position. It will be convenient to say that every ordering has a fall after the last position. In the above example, \(\sigma^{-1}\) has four falls. They occur after the second, fourth, sixth, and seventh positions. Prove that the number of rising sequences in an ordering \(\sigma\) equals the number of falls in \(\sigma^{-1}\).

A baseball player, Smith, has a batting average of .300 and in a typical game comes to bat three times. Assume that Smith's hits in a game can be considered to be a Bernoulli trials process with probability .3 for success. Find the probability that Smith gets \(0,1,2,\) and 3 hits.

Suppose that on planet Zorg a year has \(n\) days, and that the lifeforms there are equally likely to have hatched on any day of the year. We would like to estimate \(d\), which is the minimum number of lifeforms needed so that the probability of at least two sharing a birthday exceeds \(1 / 2\). (a) In Example 3.3 , it was shown that in a set of \(d\) lifeforms, the probability that no two life forms share a birthday is $$ \frac{(n)_{d}}{n^{d}} $$ where \((n)_{d}=(n)(n-1) \cdots(n-d+1)\). Thus, we would like to set this equal to \(1 / 2\) and solve for \(d\). (b) Using Stirling's Formula, show that $$ \frac{(n)_{d}}{n^{d}} \sim\left(1+\frac{d}{n-d}\right)^{n-d+1 / 2} e^{-d} $$ (c) Now take the logarithm of the right-hand expression, and use the fact that for small values of \(x\), we have $$ \log (1+x) \sim x-\frac{x^{2}}{2} $$ (We are implicitly using the fact that \(d\) is of smaller order of magnitude than \(n\). We will also use this fact in part (d).) (d) Set the expression found in part (c) equal to \(-\log (2),\) and solve for \(d\) as a function of \(n\), thereby showing that $$ d \sim \sqrt{2(\log 2) n} $$ Hint: If all three summands in the expression found in part (b) are used, one obtains a cubic equation in \(d\). If the smallest of the three terms is thrown away, one obtains a quadratic equation in \(d\). (e) Use a computer to calculate the exact values of \(d\) for various values of \(n\). Compare these values with the approximate values obtained by using the answer to part \(\mathrm{d}\) ).

Modify the program AllPermutations to count the number of permutations of \(n\) objects that have exactly \(j\) fixed points for \(j=0,1,2, \ldots, n\). Run your program for \(n=2\) to 6 . Make a conjecture for the relation between the number that have 0 fixed points and the number that have exactly 1 fixed point. A proof of the correct conjecture can be found in Wilf. \({ }^{12}\)

A poker hand is a set of 5 cards randomly chosen from a deck of 52 cards. Find the probability of a (a) royal flush (ten, jack, queen, king, ace in a single suit). (b) straight flush (five in a sequence in a single suit, but not a royal flush). (c) four of a kind (four cards of the same face value). (d) full house (one pair and one triple, each of the same face value). (e) flush (five cards in a single suit but not a straight or royal flush). (f) straight (five cards in a sequence, not all the same suit). (Note that in straights, an ace counts high or low.)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.