Question

Evaluation of ${}^{5}{{P}_{4}}$ :
(a) 720
(b) 120
(c) 60
(d) 360

Answer 1

We have to evaluate ${}^{5}{{P}_{4}}$ which is the permutation and of the form ${}^{n}{{P}_{r}}$ and we know that the expansion of ${}^{n}{{P}_{r}}$ is equal to $\dfrac{n!}{\left( n-r \right)!}$. Now, substitute n as 5 and r as 4 in this formula to get the value of ${}^{5}{{P}_{4}}$. Also, the expansion of $n!$ is equal to $n\left( n-1 \right)\left( n-2 \right).....3.2.1$ so use this expansion to solve the factorials in the formula of ${}^{n}{{P}_{r}}$.

Complete step-by-step solution:
We have to evaluate the following:
${}^{5}{{P}_{4}}$
The above expression is the permutation and of the following form:
${}^{n}{{P}_{r}}$
The expansion of the above expression in terms of factorial is as follows:
$\dfrac{n!}{\left( n-r \right)!}$
Substituting n as 5 and r as 4 in the above expression we get,
$\Rightarrow \dfrac{5!}{\left( 5-4 \right)!}$
The above expression is the expansion of ${}^{5}{{P}_{4}}$ so simplifying the above expression we get,
$\Rightarrow \dfrac{5!}{\left( 1 \right)!}$
The expansion of $5!$ is as follows:
$\begin{aligned} & =5.4.3.2.1 \\\ & =120 \\\ \end{aligned}$
And the value of $1!$ is equal to 1 so substituting the value of $5!\And 1!$ in $\dfrac{5!}{\left( 1 \right)!}$ we get,
$\begin{aligned} & =\dfrac{120}{1} \\\ & =120 \\\ \end{aligned}$
From the above, we have evaluated ${}^{5}{{P}_{4}}$ as 120.
Hence, the correct option is (b).

Note: The significance and meaning of the expression ${}^{5}{{P}_{4}}$ written in the above problem is that it means these are the possible ways to arrange 4 persons in 5 chairs. And here, all the 5 chairs are different. So, from this we can learn the concept of arrangement of n persons in r chairs or n persons in n rows.
As for permutations or arrangement of things we use ${}^{n}{{P}_{r}}$ so for combinations or selections we use ${}^{n}{{C}_{r}}$. This expression ${}^{n}{{C}_{r}}$ means the number of possible ways of selecting r items from n items in which order does not matter.