Which two numbers should be interchanged to make the given equation correct?
9 × (34 - 15) + (105 ÷ 7) + 28 + (13 × 4) = 218 (Note: Interchange should be done of entire number and not individual digits of a given number).

Given: 9 × (34 - 15) + (105 ÷7) + 28 + (13 ×4) = 218

​$$\begin{array}{|c|c|} \hline\textbf{Operation preference wise} & \textbf{Symbol} \\\hline\text{Brackets} &[],{}, () \\ \hline \text{Orders, of} & (power), √ (root) , of \\ \hline \text{Division}& ÷ \\ \hline \text{Multiplication} & × \\ \hline \text{Addition} & + \\ \hline \text{Subtraction} & - \\\hline\end{array}$$​​

Option (a): 28 and 15

9 × (34 - 28) + (105 ÷7) + 15 + (13 × 4) = 218

9 × (34 - 28) + 15 + 15 + (13 × 4) = 218

9 × 6 + 15 + 15 + 52 = 218

54 + 15 + 15 + 52 = 218

Option (b): 15 and 13

9 × (34 - 13) + (105 ÷7) + 28 + (15 × 4) = 218

9 × (34 - 13) + 15 + 28 + (15 × 4) = 218

9 × 21 + 15 + 28 + 60 = 218

181 + 15 + 28 + 60 = 218

Option (c): 34 and 28

9 × (28 - 15) + (105 ÷7) + 34 + (13 × 4) = 218

9 × (28 - 15) + 15 + 34 + (13 × 4) = 218

9 × 13 + 15 + 34 + 52 = 218

117 + 15 + 34 + 52 = 218

218 = 218

Option (d): 15 and 7

​9 × (34 - 7) + (105 ÷ 15) + 28 + (13 ×4) = 218

9 × (34 - 7) + 7 + 28 + (13 × 4) = 218

9 × 27 + 7 + 28 + 52 = 218

243 + 7 + 28 + 52 = 218

330 ≠ 218 

​It is incorrect.

Thus, the correct option is (c) 34 and 28.