Question

Find the value of x if $\dfrac{{{n}^{n}}}{n!}=\dfrac{nx}{\left( n-1 \right)!}$
A. ${{n}^{n-2}}$
B. ${{n}^{n-1}}$
C. ${{n}^{n+1}}$
D. ${{n}^{\dfrac{1}{2}}}$
E. ${{n}^{\dfrac{1}{n}}}$

Answer 1

Hint: First we will write down the meaning of factorial and how it is defined. Then, we will start by writing down the expression and expand the denominator and cancel the common terms from both sides of the equation. After that we will apply $\dfrac{{{f}^{p}}}{{{f}^{q}}}={{f}^{\left( p-q \right)}}$ and then finally get the value of $x$ .

Complete step-by-step solution:
First, let’s understand the meaning of factorial. So, basically, factorials are just products. An exclamation mark that is $!$ indicates the factorial. Factorial is a multiplication operation of natural numbers with all the natural numbers that are less than it.
The multiplication of all positive integers says $n$ that will be smaller than or equivalent to $n$ is known as the factorial. The factorial of a positive integer is represented by the symbol $n!$.
The formula to find the factorial of a number is $n!=\left( n \right)\times \left( n-1 \right)\times \left( n-2 \right)\times .............\times 3\times 2\times 1$
For an integer $n\ge 1$ , the factorial representation in terms of pi product notation is as follows:
\[n!=\prod\limits_{i=1}^{n}{i}\]
From the above formulas, the recurrence relation for the factorial of a number is defined as the product of factorial number and factorial of that number minus 1. It is given by: $n!=n.\left( n-1 \right)!$
We are given the following expression: $\dfrac{{{n}^{n}}}{n!}=\dfrac{nx}{\left( n-1 \right)!}$ ,
Now we write $n!$ as $\left( n-1 \right)!\times n$ and then we will write the given expression as follows:
$\dfrac{{{n}^{n}}}{n!}=\dfrac{nx}{\left( n-1 \right)!}\Rightarrow \dfrac{{{n}^{n}}}{\left( n-1 \right)!\times n}=\dfrac{nx}{\left( n-1 \right)!}$
We will cancel $\left( n-1 \right)!$ from both the sides of the equation and get the expression as:
$\dfrac{{{n}^{n}}}{\left( n-1 \right)!\times n}=\dfrac{nx}{\left( n-1 \right)!}\Rightarrow \dfrac{{{n}^{n}}}{n}=nx$
As we know that if $\dfrac{{{f}^{p}}}{{{f}^{q}}}={{f}^{\left( p-q \right)}}$ , therefore $\dfrac{{{n}^{n}}}{n}=nx\Rightarrow {{n}^{\left( n-1 \right)}}=nx$
Now we will take $n$ from the right hand side to the left hand side: $\begin{align}
  & \Rightarrow {{n}^{\left( n-1 \right)}}=nx\Rightarrow \dfrac{{{n}^{\left( n-1 \right)}}}{n}=x\Rightarrow {{n}^{\left( n-1-1 \right)}}=x \\
 & \Rightarrow {{n}^{\left( n-2 \right)}}=x \\
 & \Rightarrow x={{n}^{\left( n-2 \right)}} \\
\end{align}$
Therefore, the correct option is A.

Note: Note that the factorial operation is encountered in many areas of Mathematics such as algebra, permutation and combination, and mathematical analysis. Its primary use is to count “n” possible distinct objects. The student might make mistakes while subtracting the power. Remember when the bases are the same and they are in the fractional form then we subtract the power and when they are in multiplication form we add the power.