Question

Convert the following products into factorials: $\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)....\left( 2n \right)$.

Answer 1

To find the factorial we need to use a factorial that contains the elements given in the product. By multiplying the same element in both the numerator and denominator we will find the factorial. The formula to convert a product from product to factorial is:
$\dfrac{x!P}{x!}$
where $x!$ is the factorial that we have to multiply to simplify the product according to the question and $P$ is the product given in the question.

Complete step-by-step answer:
To find the factorial from the product we need two forms of factorial:
$n!=1\times 2\times 3\times ....\times (n-1)\times n$
$\left( 2n \right)!=1\times 2\times 3\times ....\times \left( n-1 \right)\times n\times \left( n+1 \right)\times \left( n+2 \right)\times ......\times (2n-1)\times 2n$
Now multiplying the $n!$ both in numerator and denominator we get:
$\Rightarrow \dfrac{n!}{n!}\times \dfrac{\left( n+1 \right)\times \left( n+2 \right)\times \left( n+3 \right)\times ....\times \left( 2n \right)}{1}$
Expanding the factorial in the above equation we get:
$\Rightarrow \dfrac{\left[ 1\times 2\times 3\times ....\times (n-1)\times n \right]\left[ \left( n+1 \right)\times \left( n+2 \right)\times \left( n+3 \right)\times ....\times \left( 2n \right) \right]}{1\times 2\times 3\times ....\times (n-1)\times n}$
$\Rightarrow \dfrac{1\times 2\times 3\times ....\times (n-1)\times n\times \left( n+1 \right)\times \left( n+2 \right)\times \left( n+3 \right)\times ....\times \left( 2n \right)}{1\times 2\times 3\times ....\times (n-1)\times n}$
Now the numerator is similar to the factorial of $2n$ and the denominator was $n!$ hence, the numerator is given as:
$\Rightarrow \dfrac{2n!}{1\times 2\times 3\times ....\times (n-1)\times n}$
$\Rightarrow \dfrac{2n!}{n!}$

$\therefore$ The factorial value of the product is $\dfrac{2n!}{n!}$.

Note: Another method to solve the question is that we know that in factorial value of $2n$is $\left( 2n \right)!=1\times 2\times 3\times ....\times \left( n-1 \right)\times n\times \left( n+1 \right)\times \left( n+2 \right)\times ......\times (2n-1)\times 2n$. Now the product given in the question is half the factorial value of $2n$. Hence, the product can also be written as:
$\Rightarrow \left( n+1 \right)\left( n+2 \right)\left( n+3 \right)....\left( 2n \right)$
$\Rightarrow \left( n+1 \right)\left( n+2 \right)\left( n+3 \right)....\left( 2n \right)=\dfrac{2n!}{1\times 2\times 3\times ....\times (n-1)\times n}$
Now changing the denominator into factorial form we get:
$\Rightarrow \left( n+1 \right)\left( n+2 \right)\left( n+3 \right)....\left( 2n \right)=\dfrac{2n!}{n!}$