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Chapter 6: Problem 30

Give a combinatorial argument to prove that $$\sum_{k=0}^{n} C(n, k)^{2}=C(2 n, n)$$

Short Answer

Expert verified
Therefore, the equality \(\sum_{k=0}^{n} C(n, k)^{2}=C(2 n, n)\) holds, as both expressions count the same combinatorial procedure (dividing 2n unique elements into two equal-sized groups) but in different ways.

Step by step solution

01

Recognize the left side of the equation

The left side of the equation \(\sum_{k=0}^{n} C(n, k)^{2}\) is essentially the sum of the squares of binomial coefficients in an row of Pascal's triangle (which represents combinations). This can be interpreted as choosing k elements from n elements , and doing this twice - hence the square.
02

Interpret the right side of the equation

The right side of the equation \(C(2 n, n)\) is picking n elements from a group of 2n elements. This can also be thought of as distributing the group of 2n elements into two groups of n each.
03

Provide Combinatorial Argument

Consider a group of 2n elements, and the task of dividing it into two groups each of size n. This can be done in \(C(2 n, n)\) ways. Alternatively, you can pick k elements from the first n elements and in the second selection, pick (n-k) from the remaining n elements, giving \(C(n, k) * C(n, n-k)\) ways. Summing this over all k from 0 to n will also give the total ways to divide the group of 2n elements into two groups of n, which was already counted as \(C(2 n, n)\). So those two quantities must be equal.

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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 a fundamental concept in combinatorics that describe the number of ways to choose a subset of items from a larger set, without regard to the order of selection. Represented mathematically as \( C(n, k) \) or \( \binom{n}{k} \), they are read as 'n choose k' and can be calculated using the formula:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \(n!\) denotes the factorial of n, meaning the product of all positive integers up to n. Binomial coefficients are intrinsic to the expansion of binomial expressions and appear as coefficients in binomial theorem expansions. They also have a close relationship with Pascal's triangle, as each number in the triangle corresponds to a binomial coefficient.
Pascal's Triangle
Pascal's triangle is a simple, yet powerful, triangular array of numbers that visually represents binomial coefficients. It is constructed by starting with a single 1 at the top, then each subsequent row starts and ends with 1, with each interior number being the sum of the two numbers directly above it from the previous row.

Here's how a small part of Pascal's Triangle looks:

1

1   1

1   2   1

1   3   3   1

1   4   6   4   1


The numbers within Pascal's triangle correspond directly to binomial coefficients, such that the nth row represents the coefficients of the expansion of \((x + y)^n\). This triangle not only provides a way to quickly find binomial coefficients but also has interesting properties related to combinatorics and probability.
Combinations
Combinations are a core concept in combinatorics, concerned with selecting items from a set where the order of selection is not important. They answer questions such as 'in how many ways can we choose k objects from a set of n objects?'. This is exactly what binomial coefficients represent; thus, combinations and binomial coefficients are often used interchangeably.

The formula to find the number of combinations is identical to that of binomial coefficients: \[ C(n, k) = \frac{n!}{k!(n-k)!} \]

In the context of the provided combinatorial proof, combinations can be visualized through Pascal's triangle or calculated directly using the formula for binomial coefficients. These concepts are interlinked, revealing the intricate structure of combinatorial mathematics.

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Most popular questions from this chapter

Answer to give an argument that proves the following result. A sequence \(a_{1}, a_{2}, \ldots, a_{n^{2}+1}\) of \(n^{2}+1\) distinct numbers contains either an increasing subsequence of length \(n+1\) or a decreasing subsequence of length \(n+1 .\) Suppose by way of contradiction that every increasing or decreasing subsequence has length \(n\) or less. Let \(b_{i}\) be the length of a longest increasing subsequence starting at \(a_{i},\) and let \(c_{i}\) be the length of a longest decreasing subsequence starting at \(a_{i} .\) Explain why \(1 \leq b_{i} \leq n\) and \(1 \leq c_{i} \leq n\).

Use induction on \(n\) to prove the Binomial Theorem.

Prove that if \(p\) ones and \(q\) zeros are placed around a circle in an arbitrary manner, where \(p, q,\) and \(k\) are positive integers satisfying \(p \geq k q,\) the arrangement must contain at least \(k\) consecutive ones.

In how many ways can three teams containing four, two, and two persons be selected from a group of eight persons?

Answer to give an argument that shows that in a group of 10 persons there are at least two such that either the difference or sum of their ages is divisible by \(16 .\) Assume that the ages are given as whole numbers. Let \(a_{1}, \ldots, a_{10}\) denote the ages. Let \(r_{i}=a_{i} \bmod 16\) and let $$ s_{i}=\left\\{\begin{array}{ll} r_{i} & \text { if } r_{i} \leq 8 \\ 16-r_{i} & \text { if } r_{i}>8 \end{array}\right. $$ Explain why \(s_{j}=s_{k}\) for some \(j \neq k\).
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