/*! 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 11 In a rural area 12 mailboxes are... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a rural area 12 mailboxes are located at a general store. a) If a newscarrier has 20 identical fliers, in how many ways can she distribute the fliers so that each mailbox gets at least one flier? b) If the mailboxes are in two rows of six each, what is the probability that a distribution from part (a) will have 10 fliers distributed to the top six boxes and 10 to the bottom six?

Short Answer

Expert verified
The number of ways the newscarrier can distribute the fliers so that each mailbox gets at least one flier is \(\binom{19}{11}\), and the probability that a distribution from has 10 fliers distributed to the top six boxes and 10 to the bottom six is \(\frac{\binom{15}{5} * \binom{15}{5}}{\binom{19}{11}}\)

Step by step solution

01

Calculate ways to distribute the fliers

First, ensure that each mailbox gets at least one flier. Hence, distribute one flier into each of 12 mailboxes. This leaves 20-12=8 fliers. Now, it is to figure out in how many ways these 8 can be distributed among 12 identical mailboxes. Since fliers are identical and can be distributed in any way among the mailboxes, this leads to the classic combinatorial problem of placing \(n+k-1\) items into \(k\) containers, which is a combinatorics problem that can be solved by calculating\(\binom{n+k-1}{k-1}\). After replacing \(n\) with 8 (remaining fliers) and \(k\) with 12 (the mailboxes), we get \(\binom{8+12-1}{12-1}\), which simplifies to \(\binom{19}{11}\).
02

Calculate the binomial coefficient

Next, compute the value of \(\binom{19}{11}\). This is a binomial coefficient problem, which can be determined by formula \(\frac{19!}{11!(19-11)!}\). The '!' symbol denotes factorial, means the product of an integer and all the integers below it (e.g., 3! is 3*2*1 = 6). Calculate this to get the number of ways the fliers can be distributed among the mailboxes.
03

Compute the probability

For part (b), we need to compute the probability of 10 fliers going into the top 6 boxes and 10 into the bottom 6. First, calculate the total number of ways to distribute 10 fliers among the 6 boxes, which is \(\binom{10+6-1}{6-1}\) = \(\binom{15}{5}\). Then, because the distributions to the top and bottom rows are independent, multiply the number of ways for the top row by the number of ways for the bottom row, which is also \(\binom{15}{5}\). This gives the total number of favourable outcomes. Now, the probability equals the number of favourable outcomes divided by the total number of outcomes (which we computed in step 1). Calculate it.

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.

Understanding the Binomial Coefficient
In combinatorics, the binomial coefficient is a key concept that helps determine the number of ways to choose a subset of items from a larger set. It is often expressed as \(\binom{n}{k}\), which represents the number of ways to select \(k\) items from \(n\) items without regard to the order of selection. This is essential when each selection or distribution should be counted only once, without repeating the same set.
The formula to calculate a binomial coefficient is \(\frac{n!}{k!(n-k)!}\). Here, "!" denotes a factorial, which is the product of all positive integers up to a certain number.
Let's consider the exercise problem: We need to distribute identical flyers to 12 mailboxes, with a requirement that each mailbox must get at least one flyer. After assigning one flyer to each mailbox, we are left to distribute the remaining flyers among the mailboxes. The number of ways to do this was expressed as \(\binom{19}{11}\). This represents the different ways to select positions for the remaining flyers, having already placed one flyer in each mailbox.
Probability in Distribution Problems
Probability is the measure of likelihood for a particular event to occur within a defined set of possibilities. In the context of distributing flyers, probability allows us to determine the likelihood of a specified distribution pattern occurring.
For example, in part (b) of our problem, calculating the probability that exactly 10 flyers will be distributed to either row of mailboxes is an important aspect.
In this case, we need the number of favorable outcomes where 10 flyers go to the top six boxes and 10 to the bottom six. This calculation uses \(\binom{15}{5}\), representing the ways to organize 10 flyers into 6 boxes (with each box receiving at least one flyer).
Multiplied by itself to cover both rows, this gives us the total ways to distribute the flyers as per the requirement. Finally, the probability of this even distribution of flyers is the ratio of these favorable outcomes to the total number of distributions possible, which we calculated in step 1 using the binomial coefficient.
Factorial Calculation Explained
Factorials are a fundamental part of calculating permutations and combinations, as well as understanding the binomial coefficient. A factorial, represented by the "!" symbol, is the product of all positive integers up to a given number.
For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow rapidly with larger numbers and are crucial in calculating the total number of possible outcomes in a combinatorial scenario.
In the given exercise problem, computing \(\binom{19}{11}\) involved calculating \(19!\), \(11!\), and \((19-11)!\), which means \(8!\). Completing these calculations allows you to find the precise value of a binomial coefficient, helping solve complex distribution and selection 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

Sergeant Bueti must distribute 40 bullets ( 20 for rifles and 20 for handguns) among four police officers so that each officer gets at least two, but no more than seven, bullets of each type. In how many ways can he do this?

a) Find the generating function for the number of ways to select 10 candy bars from large supplies of six different kinds. b) Find the generating function for the number of ways to select, with repetitions allowed, \(r\) objects from a collection of \(n\) distinct objects.

Simplify the following sum where \(n \in \mathbf{Z}^{+}:\left(\begin{array}{l}n \\\ 1\end{array}\right)+\) \(2\left(\begin{array}{l}n \\\ 2\end{array}\right)+3\left(\begin{array}{l}n \\\ 3\end{array}\right)+\cdots+n\left(\begin{array}{l}n \\ n\end{array}\right)\). (Hint: You may wish to start with the binomial theorem.)

For integers \(n, k \geq 0\) let \- \(P_{1}\) be the number of partitions of \(n\). \- \(P_{2}\) be the number of partitions of \(2 n+k\), where \(n+k\) is the greatest summand. \- \(P_{3}\) be the number of partitions of \(2 n+k\) into precisely \(n+k\) summands. Using the concept of the Ferrer's graph, prove that \(P_{1}=P_{2}\) and \(P_{2}=P_{3}\), thus concluding that the number of partitions of \(2 n+k\) into precisely \(n+k\) summands is the same for all \(k\).

Determine the generating function for the number of partitions of \(n \in \mathrm{N}\) where 1 occurs at most once, 2 occurs at most twice, 3 at most thrice, and, in general, \(k\) occurs at most \(k\) times, for every \(k \in \mathbf{Z}^{*}\).

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.