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Show that, for any integer \(n \geq 1\), $$\begin{aligned} &\text { a. }\left(\begin{array}{l} n \\ n \end{array}\right)=1 . \text { Interpret this result. }\\\ &\text { b. }\left(\begin{array}{l} n \\ 0 \end{array}\right)=1 . \text { Interpret this result. }\\\ &\text { c. }\left(\begin{array}{l} n \\ r \end{array}\right)=\left(\begin{array}{c} n \\ n-r \end{array}\right) . \text { Interpret this result. }\\\ &\mathrm{d} \cdot \sum_{i=0}^{n}\left(\begin{array}{l} n \\ i \end{array}\right)=2^{n}\left[\text { Hint: Consider the binomial expansion of }(x+y)^{n} \text { with } x=y=1 .\right] \end{aligned}$$

Short Answer

Expert verified
Each binomial coefficient describes choosing a specific number of elements, related to total combinations or subsets. Binomial expansion and symmetry simplify interrelations.

Step by step solution

01

Understanding Part (a)

Choose any integer \( n \geq 1\). We need to evaluate \( \binom{n}{n} \). The binomial coefficient \( \binom{n}{r} \) is defined as \( \frac{n!}{r!(n-r)!} \). For \( \binom{n}{n} \), substitute \( r = n \) to get \( \binom{n}{n} = \frac{n!}{n!(n-n)!} = \frac{n!}{n! \cdot 0!} = \frac{n!}{n! \cdot 1} = 1 \). Hence, \( \binom{n}{n} = 1 \).
02

Interpretation of Part (a)

This implies that there is exactly one way to choose \( n \) elements from a set of \( n \) elements. This makes intuitive sense, since if you have a set of \( n \) elements, there is exactly one way to choose all of them.
03

Understanding Part (b)

We examine \( \binom{n}{0} \). Using the formula for binomial coefficients, we get \( \binom{n}{0} = \frac{n!}{0!(n-0)!} = \frac{n!}{1 \cdot n!} = 1 \). Thus, \( \binom{n}{0} = 1 \).
04

Interpretation of Part (b)

This tells us that there is exactly one way to choose 0 elements from a set of \( n \) elements. Intuitively, choosing nothing from a set can only be done one way.
05

Understanding Part (c)

Consider \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) and \( \binom{n}{n-r} = \frac{n!}{(n-r)!r!} \). Both expressions are equal, \( \binom{n}{r} = \binom{n}{n-r} \), since the order of multiplication in the factorial doesn't change.
06

Interpretation of Part (c)

This symmetry indicates that choosing \( r \) elements from \( n \) is the same as leaving \( n-r \) elements. Essentially, selecting a group and not selecting a group are symmetric operations.
07

Understanding Part (d)

We seek to show \( \sum_{i=0}^{n} \binom{n}{i} = 2^n \). According to the binomial theorem, \((x+y)^n = \sum_{i=0}^{n} \binom{n}{i} x^{n-i} y^i\). Set \( x=1 \) and \( y=1 \) leading to \( (1+1)^n = 2^n = \sum_{i=0}^{n} \binom{n}{i} \cdot 1^{n-i} \cdot 1^i = \sum_{i=0}^{n} \binom{n}{i} \). Thus, \( \sum_{i=0}^{n} \binom{n}{i} = 2^n \).
08

Interpretation of Part (d)

This result implies that the total number of subsets of a set with \( n \) elements, including all possible sizes from 0 to \( n \), is \( 2^n \). This represents all possible combinations of elements in the power set.

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Key Concepts

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

Binomial Coefficients
Binomial coefficients are an important part of combinatorics. They are typically written as \( \binom{n}{r} \), which represents the number of ways to choose \( r \) elements from a set of \( n \) elements. The formula for calculating a binomial coefficient is \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \( ! \) denotes factorial, meaning the product of all positive integers up to a given number. For instance, \( \binom{n}{n} = 1 \) because there is exactly one way to choose all \( n \) elements from \( n \), and \( \binom{n}{0} = 1 \) because there is exactly one way to choose none.

Understanding the calculation and importance of binomial coefficients gives you a powerful tool in solving many problems involving combinations, counting, and probability. They are used to find how many different subsets can be made from a particular set, which is a frequent task in mathematical problems.
Binomial Theorem
The binomial theorem is a central aspect of algebra, providing a formula for expanding any power of a binomial in terms of coefficients. For any integer \( n \), the binomial theorem states that \((x+y)^n = \sum_{i=0}^{n} \binom{n}{i} x^{n-i} y^i\). Here is a simple breakdown of its components:
  • \( x \) and \( y \) are any numbers or variables.
  • \( n \) is the exponent, a positive integer.
  • \( \binom{n}{i} \) is the binomial coefficient for the term.
The binomial theorem is a bridge between algebra and combinatorics, helping to solve problems related to series expansion and power manipulation. In an exercise, by setting both variables to 1 like \((1+1)^n\), we encounter the interesting result \(2^n = \sum_{i=0}^{n} \binom{n}{i}\), which gives insight into subset combinations.
Subset Selection
Subset selection involves choosing elements from a set to form smaller groups or subsets. The concept is closely tied to binomial coefficients, as they effectively count how many ways you can choose a subset of a particular size.

For example, consider choosing \( r \) elements from \( n \), which can be represented by \( \binom{n}{r} \). This tells you there can be multiple ways to choose subsets of size \( r \).

To understand subset selections further, think about situations like choosing committee members from a larger group, forming teams in a sports league, or selecting a set of books from a library. Each represents a unique subset and illustrates the wide application of this concept.
Symmetry in Combinations
Symmetry in combinations is an elegant property of binomial coefficients, expressed by the formula \( \binom{n}{r} = \binom{n}{n-r} \). This means the number of ways to choose \( r \) elements from \( n \) is the same as leaving out \( n-r \) elements. In simpler terms, selecting a subset of \( r \) elements is identical to not selecting \( n-r \) elements.

Such symmetry is logical: if you have \( n \) items and you select \( r \) of them, the items you didn't choose form a distinct group of \( n-r \). Utilizing this symmetry provides simplifications in many problems, allowing for flexibility and clarity in counting methods and solving combinatorial problems easily.

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