Evaluate the following:
 ${}^{10}{P_4}$

Hint: The given $P$ is represented as the permutation of the given numbers, which is the method of counting the number of ways or arranging the numbers in the sequence.
Since the question is to find the permutation, we are going to use the method of permutation and combination methods which we will be studied on our schools to approach the given questions to find the number of ways since the number of permutations of given r-objects can be founded from among n-things is ${}^n{p_r}$where p refers to the permutation. Also, similarly, for combination, we have r-things and among n-things are ${}^n{c_r}$.
Formula used:
${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$for permutation method

Complete step by step answer:
Since from given that ${}^{10}{P_4}$ and we need to find its value
Here using the formula, we get $n = 10,r = 4$
Thus, substitute these values in the formula we get ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}} \Rightarrow {}^{10}{p_4} = \dfrac{{10!}}{{(10 - 4)!}}$
Using the subtraction operation, we know that $10 - 4 = 6$
Thus, we get ${}^{10}{p_4} = \dfrac{{10!}}{{6!}}$
Since the terms are in the form of factorial, which can be represented as in the form of $n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1$
Hence the number $10!$ can be expressed as \[10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
And the number $6!$ can be expressed as \[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
Hence substituting these values in the above we get ${}^{10}{p_4} = \dfrac{{10!}}{{6!}} \Rightarrow {}^{10}{p_4} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}$
Using the division operation, cancel the common values we have ${}^{10}{p_4} = 10 \times 9 \times 8 \times 7$ because after the number $6$ the values are the same and cancel each other.
Hence using the multiplication, we get ${}^{10}{p_4} = 5040$ which is the required answer.

Note:
If the question is about the combination like ${}^{10}{C_4}$ then we can also able to simply solve the given using the formula that ${}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Further solving we get ${}^{10}{c_4} = \dfrac{{10!}}{{4!(10 - 4)!}}$ using the factorial method we have \[{}^{10}{c_4} = \dfrac{{10!}}{{4! \times 6!}} = \dfrac{{5040}}{{24}} = 210\]
Both the formulas having an only difference in the denominator as $r!$