/*! 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 7 There are 12 men at a dance. (a)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

There are 12 men at a dance. (a) In how many ways can eight of them be selected to form a cleanup crew? (b) How many ways are there to pair off eight women at the dance with eight of these 12 men?

Short Answer

Expert verified
There are 495 ways to select the cleanup crew and 19,958,400 ways to pair off eight women with eight of these 12 men.

Step by step solution

01

Calculate the Number of Ways to Select the Cleanup Crew

The first part of the problem requires calculating the number of combinations of selecting 8 men from 12. This is given by the combination formula: \[C(n, k) = \frac{n!}{k!(n-k)!}\] where n is the total number of items, k is the number of items to select, and '!' is the factorial operator, which means the product of an integer and all the integers below it. For this problem, n=12 (total number of men) and k=8 (the number of men to select for the crew). So, the combination is \[ C(12, 8) = \frac{12!}{8!(12-8)!} = 495\]
02

Calculate the Number of Ways to Pair Off Men and Women

The second part of the problem requires calculating the number of permutations of pairing 8 women with 8 men. This is given by the permutation formula: \[P(n, r) = \frac{n!}{(n-r)!}\] where n is the total number of items to select from (in this case, the men), and r is the number of items to select (again, the men). Note that the number of women does not affect this calculation, as each man can be paired with any woman. For this problem, n=12 (total number of men) and r=8 (the number of men to pair with women). So, the permutation is \[P(12, 8) = \frac{12!}{(12-8)!} = 19,958,400\]

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 91Ó°ÊÓ!

Key Concepts

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

Combinations and Permutations
When working with sets and the task is to form subgroups, two essential concepts that come into play are combinations and permutations. Both pertain to the arrangement and selection of objects without or with considering the order, respectively.

Understanding Combinations

Combinations are used when the order of selection does not matter. For instance, if we want to form teams or groups where the position of the members isn't important, combinations are relevant. Using the given exercise as an example, forming a cleanup crew from a party of men represents a scenario where combinations would be used because the sequence in which the crew members are chosen does not affect the group's identity.

Diving Into Permutations

On the other hand, permutations are utilized when the order of selection is significant. Permutations suit situations where the arrangement is crucial, like assigning people to specific roles or seats. In the pairing of the men with the women from the dance, the order is critical since different men matched with women create distinct pairs, hence permutations apply. It’s this crucial distinguishing factor that separates permutations from combinations, and knowing when to use each one is a foundational aspect of combinatorial mathematics.
Factorial Operation
A fundamental operation in combinatorics, represented by the exclamation point (!), is the factorial. The factorial operation, denoted as 'n!', involves multiplying a series of descending natural numbers starting from 'n' down to 1. Mathematically, the expression is defined as:
\[n! = n \times (n - 1) \times (n - 2) \times \text{ ... } \times 2 \times 1\]
This operation is vital when calculating both combinations and permutations because it signifies the total number of ways to arrange 'n' distinct objects into a sequence. In the exercise presented, calculating factorials is crucial for both parts, whether selecting a subgroup or pairing individuals. The factorial encapsulates the essence of combinatorics by quantifying possibilities and outcomes.
Selecting a Subgroup from a Larger Set
Choosing a subset from a larger group is a common combinatorial task, requiring careful consideration of whether order matters (permutation) or doesn't matter (combination). The exercise shows two applications.

When forming the cleanup crew (where order doesn't matter), we use combinations to find the number of ways to select 8 out of 12 men. The formula for combinations, \(C(n, k) = \frac{n!}{k!(n-k)!}\), encapsulates this process by accounting for all possible selections without redundant counting due to order.

In pairing men to women (where order does matter), we turn to permutations, which are calculated differently because each unique arrangement counts as a distinct possibility. Here, the permutation formula \(P(n, r) = \frac{n!}{(n-r)!}\) helps us enumerate these possibilities precisely. These tools provide us with methods to manage and understand the vast potential within sets, a critical skill in various scientific and mathematical fields.

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

In how many ways can we distribute eight identical white balls into four distinct containers so that (a) no container is left empty? (b) the fourth container has an odd number of balls in it?

The expansion \(\sum_{-3}^{7} \sum_{i=1}^{4} i j\) is an example of a double sum (or double summation). Here we find that \(\sum_{j=3}^{7} \sum_{i=1}^{4} i j=\sum_{i=3}^{7}\left(\sum_{i=1}^{4} i j\right)=\sum_{j=3}^{7}(j+2 j+3 j+4 j)=\sum_{j=3}^{7} 10 j\), after we expand the inner sum(mation) for the variable \(i\). Continuing, we then expand the outer sum(mation) for the variable \(j\) and find that \(\sum_{j=3}^{7} 10 j=10 \sum_{j=3}^{7} j=10(3+4+5+6+7)=\) 250. Hence \(\sum_{j=3}^{7} \sum_{i=1}^{4} i j=250\). Determine the value of each of the following double sums. a) \(\sum_{i=1}^{4} \sum_{j=3}^{7} i j\) b) \(\sum_{i=0}^{4} \sum_{j=1}^{4}(i+j+1)\) c) \(\sum_{j=1}^{4} \sum_{i=0}^{3} i\)

A computer science professor has seven different programming books on a bookshelf. Three of the books deal with FORTRAN; the other four are concerned with BASIC. In how many ways can the professor arrange these books on the shelf (a) if there are no restrictions? (b) if the languages should alternate? (c) if all the FORTRAN books must be next to each other? (d) if all FORTRAN books must be next to each other and all BASIC books must be next to each other?

Given \(n\) distinct objects, determine in how many ways \(r\) of these objects can be arranged in a circle, where arrangements are considered the same if one can be obtained from the other by rotation.

Show that for all integers \(n, r \geq 0\), if \(n+1>r\), then $$ P(n+1, r)=\left(\frac{n+1}{n+1-r}\right) P(n, r) . $$

See all solutions

Recommended explanations on Math Textbooks

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.