Problem 7
Question
Show that \(\left(\begin{array}{c}n \\ k\end{array}\right) \cdot\left(\begin{array}{c}k \\ r\end{array}\right)=\left(\begin{array}{c}n \\\ r\end{array}\right) \cdot\left(\begin{array}{c}n-r \\\ k-r\end{array}\right)\) (for all integers \(r, k\) and \(n\) with \(r \leq k \leq n)\).
Step-by-Step Solution
Verified Answer
Using binomial coefficient properties, \( \binom{n}{k} \binom{k}{r} = \frac{n!}{r!(k-r)!(n-k)!} = \binom{n}{r} \binom{n-r}{k-r} \).
1Step 1 - Understand the Binomial Coefficient
The binomial coefficient is defined as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) where \( n!\) is the factorial of \( n\). We need to manipulate these with given criteria.
2Step 2 - Write Down the Given Expression
Write the left-hand side (LHS) of the given equation: \( \binom{n}{k} \binom{k}{r} \). This can be expanded using the definition of the binomial coefficient: \( \binom{n}{k} \binom{k}{r} = \frac{n!}{k!(n-k)!} \times \frac{k!}{r!(k-r)!} \).
3Step 3 - Simplify the LHS Expression
Simplify the expression by canceling out the \( k!\) terms: \( \frac{n!}{k!(n-k)!} \times \frac{k!}{r!(k-r)!} = \frac{n!}{r!(k-r)!(n-k)!} \).
4Step 4 - Write Down the RHS Expression
Write the right-hand side (RHS) of the equation: \( \binom{n}{r} \binom{n-r}{k-r} \). Expand using the definition: \( \binom{n}{r} \binom{n-r}{k-r} = \frac{n!}{r!(n-r)!} \times \frac{(n-r)!}{(k-r)!((n-r)-(k-r))!} \).
5Step 5 - Simplify the RHS Expression
Simplify by canceling out \( (n-r)! \) terms and simplifying the denominator: \( \frac{n!}{r!(n-r)!} \times \frac{(n-r)!}{(k-r)!(n-k)!} = \frac{n!}{r!(k-r)!(n-k)!} \).
6Step 6 - Compare the Simplified LHS and RHS
Observe that both the simplified LHS and RHS expressions are equal: \( \frac{n!}{r!(k-r)!(n-k)!} \). Therefore, \( \binom{n}{k} \binom{k}{r} = \binom{n}{r} \binom{n-r}{k-r} \).
Key Concepts
CombinatoricsFactorials in MathematicsBinomial Theorem
Combinatorics
Combinatorics is a branch of mathematics that focuses on counting, arranging, and combining objects. It plays a crucial role in solving problems where you need to figure out the number of ways certain arrangements can occur.
One key concept in combinatorics is the binomial coefficient, often denoted as \( \binom{n}{k} \). This represents the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection.
The binomial coefficient is used in a wide range of mathematical problems, including those in probability, statistics, and algebra.
One key concept in combinatorics is the binomial coefficient, often denoted as \( \binom{n}{k} \). This represents the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection.
- For example, choosing 2 fruits from a basket of 5 different fruits.
The binomial coefficient is used in a wide range of mathematical problems, including those in probability, statistics, and algebra.
Factorials in Mathematics
Factorials are a fundamental concept in mathematics, particularly useful in permutations and combinations. The factorial of any non-negative integer \( n\), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \).
For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). By definition, the factorial of zero, \( 0! \), is 1.
Factorials are key to calculating binomial coefficients:
For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). By definition, the factorial of zero, \( 0! \), is 1.
Factorials are key to calculating binomial coefficients:
- They allow us to find the number of permutations of a set.
- They are involved in formulas like the binomial theorem.
Binomial Theorem
The binomial theorem provides a quick way to expand expressions that are raised to a power. Specifically, it describes the algebraic expansion of powers of a binomial: \( (a + b)^n \).
According to the binomial theorem, this can be expressed as: \[ (a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \ ... + \binom{n}{n} a^0 b^n \]
Each term in the expansion involves a binomial coefficient, which is derived from combinations:
Understanding the binomial theorem is useful in many areas of mathematics, including calculus and statistics. It aids in simplifying complex polynomial expressions and solving algebraic equations.
According to the binomial theorem, this can be expressed as: \[ (a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \ ... + \binom{n}{n} a^0 b^n \]
Each term in the expansion involves a binomial coefficient, which is derived from combinations:
- \( \binom{n}{k} \) shows the number of ways to choose \( k \) items from \( n \) items.
- It determines the coefficient of each term in the expanded form.
Understanding the binomial theorem is useful in many areas of mathematics, including calculus and statistics. It aids in simplifying complex polynomial expressions and solving algebraic equations.
Other exercises in this chapter
Problem 6
Prove (by contraposition) that for all integers \(x\) and \(y,\) if \(x+y\) is odd, then \(x \neq y\).
View solution Problem 7
Lagrange's theorem on representation of integers as sums of squares says that every positive integer can be expressed as the sum of at most 4 squares. For examp
View solution Problem 7
Prove that if \(x\) is an odd integer, then \(x^{2}\) is of the form \(4 k+1\) for some integer \(k\).
View solution Problem 7
Prove (by contraposition) that for all real numbers \(a\) and \(b\), if \(a b\) is irrational, then \(a\) is irrational or \(b\) is irrational.
View solution