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Chapter 5: Problem 38

Use the multinomial theorem to expand \(\left(x_{1}+x_{2}+x_{3}\right)^{4}\).

Short Answer

Expert verified
The expanded form of \(\left(x_{1}+x_{2}+x_{3}\right)^{4}\) is: \(x_{1}^{4} +x_{2}^{4} + x_{3}^{4} + 4x_{1}^{3}x_{2}+4x_{1}^{3}x_{3}+4x_{2}^{3}x_{1}+ 4x_{2}^{3}x_{3}+4x_{3}^{3}x_{1}+ 4x_{3}^{3}x_{2}+ 6x_{1}^{2}x_{2}^{2}+ 6x_{1}^{2}x_{3}^{2}+ 6x_{2}^{2}x_{3}^{2}+ 12x_{1}^{2}x_{2}x_{3}+12x_{1}x_{2}^{2}x_{3}+ 12x_{1}x_{2}x_{3}^{2}\).

Step by step solution

01

Understand the multinomial theorem

First, one needs to understand the concept of the multinomial theorem. According to the theorem, \((x_{1}+x_{2}+...+x_{n})^{k}\) expands to sum of terms \(\frac{k!}{j_{1}!j_{2}!...j_{n}!}x_{1}^{j_{1}}x_{2}^{j_{2}}...x_{n}^{j_{n}}\), where \(j_{1},j_{2},...,j_{n}\) are non-negative integer and \(j_{1}+j_{2}+...+j_{n}=k\).
02

Use the multinomial theorem to expand the expression

Using the multinomial theorem, we can expand the expression \(\left(x_{1}+x_{2}+x_{3}\right)^{4}\). The expansion includes all terms \(\frac{4!}{j_{1}!j_{2}!j_{3}!}x_{1}^{j_{1}}x_{2}^{j_{2}}x_{3}^{j_{3}}\), where \(j_{1},j_{2},j_{3}\) are non-negative integer and \(j_{1}+j_{2}+j_{3}=4\).
03

Write out all possible combinations of exponents

The possible combinations of \(j_{1}, j_{2}, j_{3}\) (Exponents on \(x_{1}, x_{2}, x_{3}\)) such that their sum equals to 4 are: (4,0,0), (0,4,0), (0,0,4), (3,1,0), (3,0,1), (1,3,0), (0,1,3), (0,3,1), (2,2,0), (2,0,2), (0,2,2), (1,1,2), (1,2,1), and (2,1,1).
04

Substitute into the general term

Substitute these combinations into the general term \(\frac{4!}{j_{1}!j_{2}!j_{3}!}x_{1}^{j_{1}}x_{2}^{j_{2}}x_{3}^{j_{3}}\) and then combine the like terms to get the final answer.

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Key Concepts

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

Combinatorics
Combinatorics is the branch of mathematics that deals with counting, arrangement, and combination of elements within a set. It forms the foundation of probability, game theory, and statistics. In the context of the multinomial theorem, combinatorics helps in determining the various ways in which the terms of the polynomial can be arranged.

As an illustrative case, when expanding \( (x_1 + x_2 + x_3)^4 \) using the multinomial theorem, we count and organize the different combinations of the exponents of \( x_1 \), \( x_2 \) and \( x_3 \). The exponents need to add up to 4, since that is the power to which the trinomial is raised. It is this combinatorial aspect that leads to the formulation of multinomial coefficients.
Multinomial Coefficients
Multinomial coefficients are an extension of binomial coefficients. They arise in the expansion of a power of a multinomial 鈥� a sum of two or more terms 鈥� and signify the number of ways a specific term can be arranged. A general multinomial coefficient can be represented as \( \frac{k!}{j_1!j_2!\ldots j_n!} \), where \( k \) is the total number of items to be arranged, and \( j_1, j_2, \ldots, j_n \) are the items of different types that are to be arranged.

For example, in the multinomial expansion of \( (x_1 + x_2 + x_3)^4 \) from the exercise, \( 4! \) represents the total number of arrangements of four items, while \( j_1! \), \( j_2! \) and \( j_3! \) represent the factorial of the count of each variable, ensuring we don't overcount arrangements due to identical elements (like \( x_1 \) terms).
Polynomial Expansion
Polynomial expansion refers to the process of expressing a polynomial raised to a power as a sum of terms involving products of its coefficients and variables. The multinomial theorem provides a pattern to perform the expansion of a multinomial raised to an exponent. This involves summing over possible combinations of exponents that sum to the power, using multinomial coefficients to weigh each term.

In the exercise under discussion, the expansion of \( (x_1 + x_2 + x_3)^4 \) is the process of rewriting the expression as a polynomial where each term is a product of variables \( x_1 \), \( x_2 \) and \( x_3 \) raised to an integer exponent. The key is that the sum of these exponents in each term must be 4. The final expanded form would consist of several terms, each accompanied by a multinomial coefficient.

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Most popular questions from this chapter

Consider the partially ordered set \((X, \mid)\) on the set \(X=\\{1,2, \ldots, 12\\}\) of the first 12 positive integers, partially ordered by "is divisible by." (a) Determine a chain of largest size and a partition of \(X\) into the smallest. number of antichains. (b) Determine an antichain of largest size and a partition of \(X\) into the smallest number of chains.

Every day a student walks from her home to school, which is located 10 blocks east and 14 blocks north from home. She always takes a shortest walk of 24 blocks. (a) How many different walks are possible? (b) Suppose that four blocks east and five blocks north of her home lives her best friend, whom she meets each day on her way to school. Now how many different walks are possible? (c) Suppose, in addition, that three blocks east and six blocks north of her friend's house there is a park where the two girls stop each day to rest and play. Now how many different walks are there? (d) Stopping at a park to rest and play, the two students often get to school late. To avoid the temptation of the park, our two students decide never to pass the intersection where the park is. Now how many different walks are there?

Prove that, for all real numbers \(r\) and all integers \(k\) and \(m\), $$ \left(\begin{array}{c} r \\ m \end{array}\right)\left(\begin{array}{l} m \\ k \end{array}\right)=\left(\begin{array}{l} r \\ k \end{array}\right)\left(\begin{array}{c} r-k \\ m-k \end{array}\right) . $$

By integrating the binomial expansion, prove that, for a positive integer \(n\), $$ 1+\frac{1}{2}\left(\begin{array}{l} n \\ 1 \end{array}\right)+\frac{1}{3}\left(\begin{array}{l} n \\ 2 \end{array}\right)+\cdots+\frac{1}{n+1}\left(\begin{array}{l} n \\ n \end{array}\right)=\frac{2^{n+1}-1}{n+1} $$.

Let \(n\) and \(k\) be positive integers. Give a combinatorial proof of the identity \((5.15)\) $$ n(n+1) 2^{n-2}=\sum_{k=1}^{n} k^{2}\left(\begin{array}{l} n \\ k \end{array}\right) . $$
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