From eighty cards numbered \[1{\text{ - }}80\] , two cards are selected randomly. The probability that both the cards have the numbers divisible by $4$ is given by
 \[1){\text{ }}\dfrac{{21}}{{216}}\]
 \[2){\text{ }}\dfrac{{19}}{{316}}\]
 \[3){\text{ }}\dfrac{1}{4}\]
 \[4){\text{ }}NONE{\text{ }}OF{\text{ }}THESE\]

Hint: We have to find the required probability . We solve this question using the concept of permutation and combination . We firstly find the total possible arrangements and also the favourable outcomes using the formula of combination . The probability is given by favourable outcomes to the total possible arrangements .

Complete step-by-step answer:
Given :
There are \[80\] cards and there we need to choose $2$ cards where those cards are divisible by $4$.
Within the \[80\] cards there are \[20\] cards that are divisible by 4 that is, the numbers are \[4,8,12,16,20,24,28,32,36,40,44,48,52,60,64,68,72,76\] and \[80\] .
So ,
The total ways of selecting two cards = ${}^{80}{C_2}$
Total numbers from 1 to 80 which are divisible by \[4{\text{ }} = {\text{ }}20\]
now ,
the number of ways for favourable outcome = ${}^{20}{C_2}$
Hence , the probability that the chosen card is divisible by \[4\] = $\dfrac{{{}^{20}{C_2}}}{{{}^{80}{C_2}}}$
Also , we know that formula of combination is given as :
Using the formula and substituting the values , we get
${}^{20}{C_2}$ \[ = \] $\dfrac{{20!}}{{\left( {20 - 2} \right)! \times 2!}}$
The probability that the chosen card is divisible by $4$ \[ = \]
${}^{80}{C_2}$ \[ = \] $\dfrac{{80!}}{{\left( {80 - 2} \right)! \times 2!}}$
On solving , we get
The probability that the chosen card is divisible by $4$ \[ = \]
$\dfrac{{{}^{20}{C_2}}}{{{}^{80}{C_2}}} = \dfrac{{\dfrac{{20!}}{{\left( {20 - 2} \right)! \times 2!}}}}{{\dfrac{{80!}}{{\left( {80 - 2} \right)! \times 2!}}}}$
$ = \left( {\dfrac{{190}}{{3160}}} \right)$
$ = \dfrac{{19}}{{316}}$
Thus , the probability of selecting both divisible by $4$ is . $\dfrac{{19}}{{316}}$
Hence , the correct option is \[\left( 2 \right)\]
So, the correct answer is “Option 2”.

Note: Corresponding to each combination of \[{}^n{C_r}\] we have \[r!\] permutations, because r objects in every combinations can be rearranged in \[r!\] ways . Hence , the total number of permutations of n different things taken r at a time is \[{}^n{C_r} \times {\text{ }}r!\] . Thus \[{}^n{P_r}{\text{ }} = {\text{ }}{}^n{C_r}{\text{ }} \times {\text{ }}r!{\text{ }},{\text{ }}0 < {\text{ }}r{\text{ }} \leqslant n\;\]
Also , some formulas used :
 \[{}^n{C_1}{\text{ }} = {\text{ }}n\]
 \[{}^n{C_2} = {\text{ }}\dfrac{{{\text{ }}n\left( {n - 1} \right)}}{2}\]
 \[{}^n{C_0}{\text{ }} = {\text{ }}1\]
 \[{}^n{C_n}{\text{ }} = {\text{ }}1\]