Let's visualize the whole thing :
$int \hspace{0.1cm} {^*ptr1}=arr; \rightarrow$ Address of $10$ will be held by $ptr1$
$int \hspace{0.1cm} {^*ptr2}=arr +5; \rightarrow$ Address of $60$ i.e. $6^{th}$ element will be held by $ptr2 $ i.e. $\{base \hspace{0.1cm} address + 5^{th} \hspace{0.1cm} location\}$ $(arr+5$ numerically equals $arr + sizeof(*arr) \times 5)$
$\text{printf(``Number of elements between two pointer are: %d.",(ptr2 - ptr1));} \rightarrow$ prints "Number of elements between two pointer are: 5" on the screen
$\because[\text{Distance between the position of arr[5] & arr[0]} \hspace{0.1cm} i.e. \hspace{0.1cm} 5\hspace{0.1cm} or \hspace{0.1cm}$ $\dfrac{\text{difference between addresses }}{\text{sizeof data type pointed by that pointer variable }}$
$\qquad \qquad \qquad \qquad \qquad = \dfrac{(30-10)}{4}$ [∵ $int$ are of $4$ bytes as mentioned in question]
$\qquad \qquad \qquad \qquad \qquad = \dfrac{20}{4} = 5 ]$
$\text{printf(``Number of bytes between two pointers are: %d", (char*)ptr2 - (char*) ptr1);} \rightarrow$ prints "Number of bytes between two pointers are: 20" on the screen
[$\because ptr2$ is type-casted to $char$ pointer & size of character is $1$ byte
Number of bytes between two pointers = $\text{Number of elements between two pointers} \times \dfrac{\text{sizeof(int)}}{\text{sizeof(char)}}$
$\qquad \qquad \qquad \qquad = 5 \times \dfrac{4}{1}$ = $20$]
