Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) From the results of Section \(3.6 .3\) we can conclude that there are \(\left(\begin{array}{c}n+m-1 \\ m-1\end{array}\right)\) nonnegative integer valued solutions of the equation \(x_{1}+\cdots+x_{m}=n\) Prove this directly. (b) How many positive integer valued solutions of \(x_{1}+\cdots+x_{m}=n\) are there? Hint: Let \(y_{i}=x_{i}-1\). (c) For the Bose-Einstein distribution, compute the probability that exactly \(k\) of the \(X_{i}\) are equal to 0 .

Short Answer

Expert verified
In summary, for the given equation \(x_{1}+\cdots+x_{m}=n\), there are \(\binom{n+m-1}{m-1}\) non-negative integer-valued solutions and \(\binom{n-1}{m-1}\) positive integer-valued solutions. And for the Bose-Einstein distribution, the probability of finding exactly \(k\) zero terms is given by \(P(E) = \frac{\binom{n-1}{(m-k)-1}}{\binom{n-1}{m-1}}\).

Step by step solution

01

Understand the Equation

We are given an equation: \(x_1 + x_2 + \cdots + x_m = n\), where \(x_i\) are non-negative integers, and n and m are positive integers.
02

Use Stars and Bars Method

We can use the combinatorial method called "Stars and Bars" to count the number of non-negative integer-valued solutions for the given equation. Imagine representing the equation as a sequence of n stars and (m-1) bars. Splitting the stars by bars will give us m groups of stars, and the number of stars in each group corresponds to the value of the corresponding variable \(x_i\).
03

Calculate the Number of Combinations

Since there are n stars and (m-1) bars, we have (n+m-1) objects in total. We want to choose m-1 positions for the bars among these n+m-1 positions, so we have: \(\binom{n+m-1}{m-1}\) combinations. So, there are \(\binom{n+m-1}{m-1}\) non-negative integer-valued solutions for the given equation. #b) Find the number of positive integer-valued solutions.#
04

Apply the Hint and Define New Variables

The hint suggests to use the transformation: \(y_i = x_i - 1\). Now, let's apply this substitution to the original equation: \((y_1 + 1) + (y_2 + 1) + \cdots + (y_m + 1) = n\).
05

Simplify the New Equation

In this new equation, the variables \(y_i\) can be considered as non-negative integers instead of positive ones: \(y_1 + y_2 + \cdots + y_m = n - m\).
06

Apply the Result of Part (a)

Since we found the number of non-negative integer-valued solutions in part (a), we can use the same result for the new equation: There are \(\binom{(n-m)+(m-1)}{m-1}\) positive integer-valued solutions. So, there are \(\binom{n-1}{m-1}\) positive integer-valued solutions for the given equation. #c) Compute the probability of finding exactly k zero terms.#
07

Define the Probability

Let's denote the event of finding exactly k zero terms in the given equation as E. We need to find \(P(E)\), the probability of E.
08

Apply the Bose-Einstein Distribution Formula

The Bose-Einstein distribution formula for this problem can be described as: \(P(E) = \frac{\text{Number of positive integer-valued solutions with k zero terms}}{\text{Total Number of integer-valued solutions}}\).
09

Calculate the Number of Positive Integer-valued Solutions with k Zero Terms

If exactly k terms are equal to 0, that means we need to find the number of positive integer-valued solutions for the equation with (m-k) variables instead of m variables: \(\binom{n-1}{(m-k)-1}\)
10

Calculate the Probability

Now, we can find the probability of E: \(P(E) = \frac{\binom{n-1}{(m-k)-1}}{\binom{n-1}{m-1}}\).

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 Vaia!

Key Concepts

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

Stars and Bars Theorem
The Stars and Bars Theorem is a fundamental principle in combinatorial mathematics used to solve problems involving non-negative integer solutions to equations. To visualize the theorem, consider representing an equation as a sequence of stars (which denote the items to be distributed) and bars (which separate the groups receiving the items).

For example, if you want to distribute n identical objects into m distinct boxes, you can think of n stars and (m-1) bars. The bars serve to divide the stars into m groups. The crucial realization is that solving for the number of non-negative integer valued solutions is equivalent to finding all possible ways to place the (m-1) bars between the stars or at the ends. Therefore, the total number of ways to do this is given by the binomial coefficient \binom{n+m-1}{m-1}\, which represents choosing (m-1) positions for the bars out of (n+m-1) possible placements.

Applying this to an equation such as \(x_1 + x_2 + \text{...}\ + x_m = n\) showcases the theorem's utility: it simplifies the complicated task of counting the number of solutions into a simple combinatorial problem. To ensure that students understand this concept, encourage them to draw representations with stars and bars for small values of n and m. This hands-on approach clarifies the abstract concept into something tangible and visually understandable.
Bose-Einstein Distribution
The Bose-Einstein Distribution is a statistical distribution that applies to particles known as bosons. In the context of combinatorial probability, the Bose-Einstein Distribution can be used to model scenarios where identical objects are distributed into distinguishable boxes with any number of objects per box permissible. In such cases, the objects can be viewed similarly to bosons that don't obey the Pauli exclusion principle and can occupy the same state.

When you calculate the probability of a specific event in this distribution, you are often counting the ways certain conditions can be met within the distribution. For instance, finding the probability that exactly k of the terms (which can be seen as boxes in this analogy) are equal to zero is akin to calculating the likelihood that k boxes contain no bosons. To grasp this principle, students are encouraged to consider simpler cases, such as finding the probability of having exactly one zero term, and then expanding to more complex scenarios. This incremental approach helps them understand the general formula for these probabilities by building up from more familiar instances.
Positive Integer Solutions
Finding Positive Integer Solutions to an equation involves determining all possible distributions of a whole number n across m variables where each variable is at least 1. Unlike non-negative integer solutions, the allowance of zero is not permitted here, adding a layer of complexity to the problem.

To find the number of positive solutions for an equation like \(x_1 + x_2 + \text{...}\ + x_m = n\), one commonly used strategy involves a simple transformation. By defining new variables such as \(y_i = x_i - 1\), the equation transforms into one where non-negative solutions are allowed, and one can then apply the Stars and Bars Theorem. The transformation effectively reduces the original equation by m since one is subtracted from each variable. This leads to the new equation \(y_1 + y_2 + \text{...}\ + y_m = n - m\), and the Stars and Bars Theorem gives us the count of solutions as \binom{n-1}{m-1}\.

By teaching students to perform such variable transformations, they can leverage their understanding of non-negative solution counting to tackle problems that require positive solutions only. Practice with variable substitution may further strengthen their grasp of the concept.
Combinatorial Methods
Combinatorial Methods refer to techniques used to count, arrange, or combine items in various ways without having to explicitly enumerate all possibilities. These methods are crucial in solving problems related to probability, optimization, and decision-making. The Stars and Bars Theorem, as mentioned earlier, is one such method.

When applying combinatorial methods, one often employs factorials, permutations, combinations (binomial coefficients), and other counting principles such as the Rule of Sum and Rule of Product. For instance, combinations are used when the order does not matter, while permutations are used when the order is significant. Students are usually introduced to combinatorial methods through problems involving simpler concepts like choosing committees or seating arrangements. Eventually, they progress to more complex applications like those found in combinatorial proofs, generating functions, and graph theory.

Since combinatorial methods can become quite abstract, providing real-world examples or gamified activities can greatly aid in understanding. Examples such as spreading out a set of cards into piles or distributing colored balls into different boxes can make these concepts more concrete and relatable for students, ultimately promoting a deeper comprehension.

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

Suppose each new coupon collected is, independent of the past, a type \(i\) coupon with probability \(p_{i} .\) A total of \(n\) coupons is to be collected. Let \(A_{i}\) be the event that there is at least one type \(i\) in this set. For \(i \neq j\), compute \(P\left(A_{i} A_{j}\right)\) by (a) conditioning on \(N_{i}\), the number of type \(i\) coupons in the set of \(n\) coupons; (b) conditioning on \(F_{i}\), the first time a type \(i\) coupon is collected; (c) using the identity \(P\left(A_{i} \cup A_{j}\right)=P\left(A_{i}\right)+P\left(A_{j}\right)-P\left(A_{i} A_{j}\right)\).

A coin having probability \(p\) of coming up heads is successively flipped until two of the most recent three flips are heads. Let \(N\) denote the number of flips. (Note that if the first two flips are heads, then \(N=2 .\) ) Find \(E[N]\).

Two players alternate flipping a coin that comes up heads with probability \(p\). The first one to obtain a head is declared the winner. We are interested in the probability that the first player to flip is the winner. Before determining this probability, which we will call \(f(p)\), answer the following questions. (a) Do you think that \(f(p)\) is a monotone function of \(p ?\) If so, is it increasing or decreasing? (b) What do you think is the value of \(\lim _{p \rightarrow 1} f(p) ?\) (c) What do you think is the value of \(\lim _{p \rightarrow 0} f(p) ?\) (d) Find \(f(p)\).

Suppose that we want to predict the value of a random variable \(X\) by using one of the predictors \(Y_{1}, \ldots, Y_{n}\), each of which satisfies \(E\left[Y_{i} \mid X\right]=X .\) Show that the predictor \(Y_{i}\) that minimizes \(E\left[\left(Y_{i}-X\right)^{2}\right]\) is the one whose variance is smallest. Hint: Compute \(\operatorname{Var}\left(Y_{i}\right)\) by using the conditional variance formula.

Suppose \(p(x, y, z)\), the joint probability mass function of the random variables \(X\), \(Y\), and \(Z\), is given by $$ \begin{array}{ll} p(1,1,1)=\frac{1}{8}, & p(2,1,1)=\frac{1}{4} \\ p(1,1,2)=\frac{1}{8}, & p(2,1,2)=\frac{3}{16} \\ p(1,2,1)=\frac{1}{16}, & p(2,2,1)=0 \\ p(1,2,2)=0, & p(2,2,2)=\frac{1}{4} \end{array} $$ \text { What is } E[X \mid Y=2] ? \text { What is } E[X \mid Y=2, Z=1] ?

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.

Sign-up for free