Question

How do you simplify $^5P_2$ ?

Answer 1

Hint: Given question is from permutations and combinations. We have to find the value of .This can be solved directly by using the formula for permutations \[ (^nP_r) = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] where n is the total number of whatever is being arranged or selected and r is the number of objects getting arranged or selected.
This is just to get an idea of different possible combinations. So let’s solve it!

Complete step-by-step answer:
Given that $^5P_2$
Using the formula
\[ (^nP_r) = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
 \[ \Rightarrow (^5P_2) = \dfrac{{5!}}{{\left( {5 - 2} \right)!}}\]
On soling we get,
 \[ \Rightarrow (^5P_2) = \dfrac{{5!}}{{3!}}\]
Now let’s write the values of factorials there,
 \[ \Rightarrow (^5P_2) = \dfrac{{120}}{6}\]
On dividing we get,
 \[ \Rightarrow (^5P_2) = 20\]
This is our answer.
So, the correct answer is “20”.

Note: Permutations mainly involve the possible number of arrangements of r objects chosen from n different objects. Factorial is the product of numbers below that number upto 1. For example here \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\] . Starting from that number and taking all numbers below it upto 1.
Note that \[{\text{n}}{{\text{P}}_1} = n\] and \[{\text{n}}{{\text{P}}_n} = n!\] because \[0! = 1\]