Problem 78 - Algebra and Trigonometry

Question

Why is \(\left(\begin{array}{c}{n} \\ {r}\end{array}\right)\) the Same as \(C(n, r) ?\) This exercise explains why the binomial coefficients \(\left(\begin{array}{c}{n} \\ {r}\end{array}\right)\) that appear in the expansion of \((x+y)^{n}\) are the same as \(C(n, r),\) the number of ways of choosing \(r\) objects from \(n\) objects. First, note that expanding a binomial using only the Distributive Property gives $$\begin{aligned}(x+y)^{2} &=(x+y)(x+y) \\ &=(x+y) x+(x+y) y \\ &=x x+x y+y x+y y \end{aligned}$$ $$\begin{aligned}(x+y)^{3}=&(x+y)(x x+x y+y x+y y) \\\=& x x+x x y+x y x+x y y+y x x \\ &+y x y+y y x+y y y \end{aligned}$$ (a) Expand \((x+y)^{5}\) using only the Distributive Property. (b) Write all the terms that represent \(x^{2} y^{3}\) together. These are all the terms that contain two \(x^{\prime}\) s and three \(y^{\prime}\) s. (c) Note that two \(x^{\prime}\) s appear in all possible positions. Conclude that the number of terms that represent \(x^{2} y^{3}\) is \(C(5,2) .\) (d) In general, explain why \(\left(\begin{array}{c}{n} \\\ {r}\end{array}\right)\) in the Binomial Theorem is the same as \(C(n, r) .\)

Step-by-Step Solution

Answer

The binomial coefficient \\left(\begin{array}{c}{n} \\ {r}\end{array}\right)\\ equals \\C(n, r)\\ because it represents choosing r x's from n positions in the binomial expansion.

Step 1: Understanding the Problem

We need to expand \(x+y\)^5 using only the Distributive Property. This requires expanding every possible combination of choosing terms from each factor and then identifying particular combinations.

Step 2: Expand \\(x+y\\)^5

Start by expanding \(x+y\)^5 step-by-step using the Distributive Property: \(x+y\)^{5} = (x+y)(x+y)(x+y)(x+y)(x+y). Each step, multiply the expression by another \(x+y\), and expand it fully until you reach the final expanded form. Write down all individual terms you obtain.

Step 3: Identify x^2y^3 Terms

Within the expansion of \(x+y\)^5, locate all the terms that are of the form \x^2 y^3\. Each of these terms will contain exactly two x's and three y's, regardless of order, such as \xxyyy, xyxyy, xyxyy, ...\, etc.

Step 4: Count the Terms

Count the number of distinct terms that match the form \(x^2 y^3\). Recognize that arranging two x's in five total positions corresponds to choosing 2 positions from 5, which is given by \C(5, 2)\.

Step 5: Generalize with the Binomial Coefficient

Recognize that in the expansion of \(x+y\)^n, terms of the form \x^r y^{(n-r)}\ are generated by choosing r positions for x out of n total positions. This choice correlates with the binomial coefficient \left(\begin{array}{c}{n} \ {r}\end{array}\right)\ and is the same as \C(n, r)\.

Key Concepts

Binomial Theorem

The Binomial Theorem is an essential concept in algebra that allows us to expand expressions raised to a power, like \((x + y)^n\). The formula is given by:\[(x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r\]where \(\binom{n}{r}\) represents the binomial coefficients, which tell us how many different ways we can pick \(r\) items from a total of \(n\) items. These coefficients are the same ones we see in Pascal's Triangle.

• Each term in the expansion corresponds to a different way of distributing powers between \(x\) and \(y\).
• The sum of the exponents in each term \'s expansion is always equal to \(n\).
• The Binomial Theorem simplifies the process of expanding large powers by providing a direct formula instead of expanding step by step.

Understanding how the binomial coefficients play a role in the theorem connects algebraic expansion with combinatorial counting.

Combinatorics

Combinatorics is the branch of mathematics focused on counting, arrangement, and the combination of objects. It plays a crucial role when we talk about binomial coefficients in the context of the Binomial Theorem. Consider a scenario where you need to determine how many different ways you can choose \(r\) objects from a group of \(n\) objects; this is expressed as:\[C(n, r) = \binom{n}{r}\]

• These calculations help us figure out how many distinct terms appear in the expansion of a binomial expression like \((x+y)^n\).
• The concept ensures that when multiplying binomials, every possible combination of \(x\)'s and \(y\)'s is included, revealing all potential terms like \(x^2 y^3\).
• Understanding these combinations allows us to see patterns and relationships between elements and understand how elements are distributed in groups.

This connection between combinatorics and binomial coefficients shows the power of counting methods in solving algebraic expressions.

Distributive Property

The Distributive Property is a fundamental algebraic principle used to expand expressions. It states that:\[a(b + c) = ab + ac\]For binomial expansions, this property enables us to gradually break down and simplify expressions like \((x + y)^n\) by continually distributing each term. For instance:

• Applying the property to \((x+y)^2\) expands it to \(xx + xy + yx + yy\).
• Continuing this pattern allows us to expand powers such as \((x+y)^5\), systematically exploring every combination of the terms.
• The importance of this property lies in its ability to manage and simplify variables across multiple operations, revealing all possible terms and arrangements.

In the context of binomial expansions, the Distributive Property helps order combinations, making it easier to identify specific terms like \(x^2 y^3\), and provides a step-by-step approach until all terms are clear and accounted for.