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Prove the hockeystick identity $$ \sum_{k=0}^{r}\left(\begin{array}{c}{n+k} \\\ {k}\end{array}\right)=\left(\begin{array}{c}{n+r+1} \\ {r}\end{array}\right) $$ whenever \(n\) and \(r\) are positive integers, a) using a combinatorial argument. b) using Pascal's identity.

Short Answer

Expert verified
Use combinatorial arguments to show both sides count the same thing. Apply Pascal's identity to show the equivalence step by step.

Step by step solution

01

- Define the Identity

The hockeystick identity states that \[\sum_{k=0}^{r}\left(\begin{array}{c}{n+k} \ {k}\end{array}\right)=\left(\begin{array}{c}{n+r+1} \ {r}\end{array}\right)\]for positive integers \(n\) and \(r\).
02

- Combinatorial Argument

Consider a set with \(n+r+1\) elements. We want to count the number of ways to choose \(r\) elements out of this set. One way to do this is by grouping the elements into subsets.First, choose \(k\) elements from the first \(n+k\) elements and then choose the remaining \(r-k\) elements from the remaining \(n+r+1-(n+k)\) elements.Summing over all possible \(k\) from 0 to \(r\) gives our left hand side.Thus, \[\sum_{k=0}^{r}\left(\begin{array}{c}{n+k} \ {k}\end{array}\right)\]counts the same thing as \[\left(\begin{array}{c}{n+r+1} \ {r}\end{array}\right)\].
03

- Using Pascal's Identity

Pascal's identity states that \[\left(\begin{array}{c}{n+1} \ {k}\end{array}\right)= \left(\begin{array}{c}{n} \ {k-1}\end{array}\right)+\left(\begin{array}{c}{n} \ {k}\end{array}\right)\]. Using this identity, we can rewrite each term in the sum:\[\left(\begin{array}{c}{n+k} \ {k}\end{array}\right)=\left(\begin{array}{c}{n+1+k-1} \ {k}\end{array}\right)=\left(\begin{array}{c}{n+k-1} \ {k-1}\end{array}\right)+ \left(\begin{array}{c}{n+k-1} \ {k}\end{array}\right)\].Notice this forms a pattern that sums up to \[\left(\begin{array}{c}{n+r+1} \ {r}\end{array}\right)\].

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Key Concepts

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

Combinatorial Argument
A combinatorial argument provides a visual and conceptual way to understand and prove the hockeystick identity. Suppose we have a set of \(n+r+1\) elements. Our goal is to choose \(r\) elements from this set. Think about how we can break this down into smaller steps.

One method is to first choose \(k\) elements from the first \(n+k\) elements. This selection can be represented using the binomial coefficient \(\binom{n+k}{k}\). Once you choose these \(k\) elements, the remaining \(r-k\) elements must be chosen from the remaining \((n+r+1)-(n+k) = r+1-k\) elements.

As we sum over all possible values of \(k\) from 0 to \(r\), the different ways to pick the elements are exhaustively counted. Thus, the left hand side of the identity sums these different combinations: \(\text{\textbackslash sum}\textbackslash _{k=0}\textbackslash ^{r}\textbackslash \binom{n+k}{k}\).

This sum, by construction, simply counts all the ways to choose \(r\) elements from the entire set of \(n+r+1\) elements. Therefore, it is the same as the binomial coefficient representing this choice directly: \(\binom{n+r+1}{r}\). This completes the proof using a combinatorial argument.
Pascal's Identity
Pascal's identity is a fundamental property of binomial coefficients, which states that \(\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}\). We can use this identity to prove the hockeystick identity.

To begin, let’s rewrite each term in the sum on the left-hand side of the hockeystick identity using Pascal's identity: \(\binom{n+k}{k} = \binom{n+k-1}{k-1} + \binom{n+k-1}{k}\).

By applying Pascal's identity recursively, we begin to see a pattern. Summing these identities over all values of \(k\) from 0 to \(r\), each pair of terms \( \binom{n+k-1}{k-1} + \binom{n+k-1}{k}\) will add up, ultimately collapsing into a single term: \(\binom{n+r+1}{r}\).

Therefore, through the recursive application of Pascal's identity, we confirm that the left-hand side of the hockeystick identity equals the right-hand side.
Binomial Coefficients
Binomial coefficients are a crucial concept in combinatorics, often displayed as \(\binom{n}{k}\) and read as 'n choose k'. They represent the number of ways to choose \(k\) elements from \(n\) elements without regard to the order of selection.

The formula for calculating a binomial coefficient is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). This expression arises from the ways of arranging \(n\) elements and selecting \(k\) of them.

Binomial coefficients also appear in the binomial theorem, which provides a way to expand expressions of the form \((a + b)^n\). They are instrumental in various combinatorial arguments, including the proof of the hockeystick identity.

Understanding these coefficients and their properties, such as symmetry (\binom{n}{k} = \binom{n}{n-k})) and recursive relationships (Pascal's identity), can greatly aid in solving complex combinatorial problems.

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Most popular questions from this chapter

In this exercise we will count the number of paths in the \(x y\) plane between the origin \((0,0)\) and point \((m, n),\) where \(m\) and \(n\) are nonnegative integers, such that each path is made up of a series of steps, where each step is a move one unit to the right or a move one unit upward. (No moves to the left or downward are allowed.) Two such paths from \((0,0)\) to \((5,3)\) are illustrated here. a) Show that each path of the type described can be represented by a bit string consisting of \(m\) 0s and \(n\) ls, where a 0 represents a move one unit to the right and a 1 represents a move one unit upward. b) Conclude from part (a) that there are \(\left(\begin{array}{c}{m+n} \\\ {n}\end{array}\right)\) paths of the desired type.

Give a formula for the coefficient of \(x^{k}\) in the expansion of \((x+1 / x)^{100},\) where \(k\) is an integer.

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a) Suppose that a popular style of running shoe is available for both men and women. The woman’s shoe comes in sizes 6, 7, 8, and 9, and the man’s shoe comes in sizes 8, 9, 10, 11, and 12. The man’s shoe comes in white and black, while the woman’s shoe comes in white, red, and black. Use a tree diagram to determine the number of different shoes that a store has to stock to have at least one pair of this type of running shoe for all available sizes and colors for both men and women. b) Answer the question in part (a) using counting rules.

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