# Factorial coefficient simplification

• physicsjock
In summary: Science ExpertIn summary, there is a way to simplify the given series coefficients without using the ...'s. By using the factorial notation and cancelling out common terms, we can represent the expressions as fractions with simpler terms. This can help in understanding the patterns and simplifying the coefficients.
physicsjock
Hey,

I've been trying to find a way to simplify these two series coefficients,

\begin{align} & {{a}_{4n}}={{a}_{o}}\frac{2!6!...(n-6)!(n-2)}{4!8!...(n-4)!n!} \\ & {{a}_{4n+1}}={{a}_{1}}\frac{3!7!...(n-6)!(n-2)}{5!9!...(n-4)!n!} \\ & E.g. \\ & {{a}_{12}}={{a}_{o}}\frac{2!6!10!}{4!8!12!}\,= \frac{{{a}_{o}}}{3\cdot 4\cdot 7\cdot 8\cdot 11\cdot 12}\,\,\,\,\,\,and \\ & {{a}_{13}}={{a}_{1}}\frac{3!7!11!}{5!9!13!}\,=\,\,\frac{{{a}_{1}}}{4\cdot 5\cdot 8\cdot 9\cdot 12\cdot 13} \\ \end{align}

This is the best way I could find but I was wondering if there were any ways to avoid the use of ...'s

Is there any way to simplify these coefficients?

!

Hi there,

I can definitely help you with simplifying these series coefficients. It's great that you're trying to find a simpler way to represent them without using the ...'s. Let's take a closer look at the expressions you have provided:

\begin{align}
& {{a}_{4n}}={{a}_{o}}\frac{2!6!...(n-6)!(n-2)}{4!8!...(n-4)!n!} \\
& {{a}_{4n+1}}={{a}_{1}}\frac{3!7!...(n-6)!(n-2)}{5!9!...(n-4)!n!} \\
\end{align}

One way to simplify these coefficients is by using the factorial notation, denoted by the exclamation mark (!). For example, instead of writing 2!6!, we can write it as 2x1x6x5x4x3x2x1. This makes it easier to see the pattern and simplify the expressions.

Now, let's take a look at the first expression. We can rewrite it as:

\begin{align}
{{a}_{4n}}={{a}_{o}}\frac{(n-2)!}{4!(n-4)!n!}
\end{align}

Notice that the terms in the numerator and denominator are very similar. We can further simplify this by cancelling out the common terms, resulting in:

\begin{align}
{{a}_{4n}}={{a}_{o}}\frac{1}{4n(4n-1)(4n-2)(4n-3)}
\end{align}

Similarly, we can simplify the second expression as:

\begin{align}
{{a}_{4n+1}}={{a}_{1}}\frac{(n-2)!}{5!(n-4)!n!}={{a}_{1}}\frac{1}{5n(5n-1)(5n-2)(5n-3)}
\end{align}

This way, we have avoided using the ...'s and simplified the expressions to a great extent. I hope this helps you in your research. Keep up the good work!

## What is a factorial coefficient?

A factorial coefficient is a mathematical term used to describe the number of ways to choose a group of objects from a larger set. It is represented by the symbol "n choose k" and is calculated as n! / (k! * (n-k)!) where n is the total number of objects and k is the number of objects being chosen.

## How do you simplify a factorial coefficient?

To simplify a factorial coefficient, you can use the formula n! / (k! * (n-k)!). Additionally, if k is a small number, you can easily calculate the factorial values and simplify the expression using basic arithmetic operations.

## What is the purpose of simplifying a factorial coefficient?

Simplifying a factorial coefficient can help to reduce complex expressions and make them easier to understand and work with. It is often used in combinatorics and probability calculations.

## What are some common applications of factorial coefficient simplification?

Factorial coefficient simplification is commonly used in various fields such as statistics, genetics, and computer science. It is used to calculate the number of possible outcomes in a given scenario, such as the number of ways to arrange a deck of cards or the number of genetic combinations in offspring.

## Are there any limitations to using factorial coefficient simplification?

While factorial coefficient simplification is a useful tool, it is limited to situations where the objects being chosen are distinct and order does not matter. It also assumes that all objects have an equal chance of being chosen, which may not always be the case in real-world scenarios.

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