Answer
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Hint: For solving the factorial given in the above question, we need to use the definition of the factorial of a number. The factorial of an integer is equal to the product of the integer and all of the integers below it upto one. So in general the factorial of a number n can be given as $n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).......3\cdot 2\cdot 1$. Therefore, the given factorial of $20$ will be equal to the product of all the integers from one to twenty. On solving this product, we will obtain the required value for the factorial of $20$.
Complete step-by-step solution:
According to the question, we have been given the factorial of $20$ and are asked to find its value.
We know that the factorial of an integer is equal to the product of the integer and all of the integers below it upto one. Therefore, the general form for the factorial of an integer n can be written as
$\Rightarrow n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).......3\cdot 2\cdot 1$
On substituting $n=20$ in the above equation, we get
$\begin{align}
& \Rightarrow 20!=20\left( 20-1 \right)\left( 20-2 \right)\left( 20-3 \right).......3\cdot 2\cdot 1 \\
& \Rightarrow 20!=20\cdot 19\cdot 18\cdot 17.......3\cdot 2\cdot 1 \\
\end{align}$
On solving the above product we finally get
$\Rightarrow 20!=2432902008176640000$
Hence, the factorial of $20$ has been finally found to be equal to $2432902008176640000$.
Note: We must note that the factorial is defined only for non-negative integers. The factorial of a negative integer cannot be found out. Also, there does not exist any special properties for the factorial of a number. So there is no option other than to calculate the long product in the above solution.
Complete step-by-step solution:
According to the question, we have been given the factorial of $20$ and are asked to find its value.
We know that the factorial of an integer is equal to the product of the integer and all of the integers below it upto one. Therefore, the general form for the factorial of an integer n can be written as
$\Rightarrow n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).......3\cdot 2\cdot 1$
On substituting $n=20$ in the above equation, we get
$\begin{align}
& \Rightarrow 20!=20\left( 20-1 \right)\left( 20-2 \right)\left( 20-3 \right).......3\cdot 2\cdot 1 \\
& \Rightarrow 20!=20\cdot 19\cdot 18\cdot 17.......3\cdot 2\cdot 1 \\
\end{align}$
On solving the above product we finally get
$\Rightarrow 20!=2432902008176640000$
Hence, the factorial of $20$ has been finally found to be equal to $2432902008176640000$.
Note: We must note that the factorial is defined only for non-negative integers. The factorial of a negative integer cannot be found out. Also, there does not exist any special properties for the factorial of a number. So there is no option other than to calculate the long product in the above solution.
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