**METHOD 1.**

**$\rightarrow$Total no. of attempts combination possible for a student ** (= **no. of holes =m)**

= For each question he can choose any 1 of the 4 options and there are 5 such questions

$ = 4*4*4*4*4$

$ = 1024.$

**$\rightarrow$** We need to find no. of students (**= no. of pigeons = n**)

**$\rightarrow$** It is given that at least $4$ answer sheets will be identical (= **no. of pigeons that share the same hole should be at least 4.**)

According to **pigeon hole principle** ,

$\left \lfloor \frac{(n-1)}{m} \right \rfloor+1=4$

$\Rightarrow$ $\left \lfloor \frac{(n-1)}{1024} \right \rfloor =3$

$\Rightarrow n-1 =3 *1024=3072$

$\Rightarrow n=3073$

**METHOD 2.**

In total there are $4*4*4*4*4=1024$ combination of options selection possible and a student will choose any one combination out of it.

Questions is asking how many students should be there so that atleast $4$ of them selects the same combination.

suppose $1024$ students comes and pick $1024$ unique combinations.

then another $1024$ students come and pick $1024$ unique combinations

then another $1024$ students come and pick $1024$ unique combinations.

So uptil here each combination is selected by $3$ students and total students are $1024+1024+1024 = 3072.$

Now if a student comes and select a combination i.e. select any one of $1024$ combinations then we will have $1$ combination that is selected by $4$ student.

so minimum students required $= 3072+1 = 3073$