# Permutation and counting problem

• cientifiquito
In summary, the problem asks to show that (-1/2)r is equal to (-1)rr!2-2r(2r take r). One way to approach this is by starting from the definition of (x)r and plugging in -1/2, and from the given expression for (-1/2)r and finding a way to write it in terms of a product of even integers.
cientifiquito

## Homework Statement

Let r be a positive integer. For any number x, let
(x)r = x(x-1)(x-2)...(x-r+1)

Show that
(-1/2)r = (-1)rr!2-2r(2r take r)

## Homework Equations

by "2r take r" I mean what is usually denoted by (n / r) (written like a fraction but without the bar) and is calculated as: n!/(r!(n-r)!)

## The Attempt at a Solution

If I start from the definition of (x)r, plugging in -1/2, I get as far as:

(-1)r(-1/2)r(1)(1+2)(1+4)(1+6)...(1+2r-2)
i.e.,

(-1)r(-1/2)r(1)(1+2)(1+4)(1+6)...(2r - 1)

And if I start from what I'm supposed to be showing that (-1/2)r is equal to, I can get to

(-1)r(-1/2)r[(2r)!/(r!(2r-r)!)]
i.e.,
(-1)r(-1/2)r[(r+1)(r+2)...(2r)]

but obviously I'm not seeing the connection between the two

You've got a product of odd integers 1.3.5...(2r-1); the problem is how you write that compactly.
The trick here is that we can compactly write a product of even integers 2.4.6...(2r), and then you can write 1.3.5...(2r-1)=[1.2.3.4...(2r)]/[2.4.6...(2r)].

I see now, thanks henry_m

## 1. What is the difference between permutation and combination?

Permutation refers to the arrangement of elements in a specific order, while combination refers to the selection of elements without considering the order. In other words, permutation deals with the number of ways to arrange a set of objects, while combination deals with the number of ways to select a subset of objects.

## 2. How do I calculate the number of permutations?

The formula for calculating the number of permutations is n! / (n-r)!, where n is the total number of objects and r is the number of objects being selected. This formula is used when order matters, and there are no repeats in the set of objects.

## 3. Can I use permutations to solve real-life problems?

Yes, permutations can be used to solve real-life problems such as arranging books on a shelf, seating arrangements at a dinner table, or creating unique passwords. It is a useful concept in fields such as computer science, statistics, and economics.

## 4. What are some common misconceptions about permutations and counting problems?

One common misconception is that the order of elements does not matter in permutations. Another misconception is that repetition is not allowed in permutations, which is not always the case. It is important to understand the specific problem and the rules before applying the concept of permutations.

## 5. How can I practice and improve my skills in solving permutation and counting problems?

There are many resources available online, such as practice problems and tutorials, that can help improve your skills in solving permutation and counting problems. It is also helpful to familiarize yourself with the different types of counting problems, such as permutations, combinations, and variations, and practice using the appropriate formulas for each type.

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