Which two numbers should be interchanged to make the given equation correct?
50 × 3 + 12 × 5 - (66 ÷ 2) + 11 × 4 - 22 = 160 (Note: Interchange should be done of entire number and not individual digits of a given number)

Given: 50 × 3 + 12 × 5 – (66 ÷ 2) + 11 × 4 – 22 = 160

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

Option (a): 3 and 2

50 × 2 + 12 × 5 – (66 ÷ 3) + 11 × 4 – 22 = 160

50 × 2 + 12 × 5 – 22 + 11 × 4 – 22 = 160

100 + 60 - 22 + 44 -22 = 160

160 + 44 - 22 - 22 = 160

204 - 44 = 160

Option (b): 5 and 3

50 × 5 + 12 × 3 – (66 ÷ 2) + 11 × 4 – 22 = 160

50 × 5 + 12 × 3 – 33 + 11 × 4 – 22 = 160

250 + 36 - 33 + 44 - 22 = 160

250 + 36 + 44 - 33 - 22 = 160

330 - 33 - 22 = 160

330 - 55 = 160 

Option (c): 22 and 2

50 × 3 + 12 × 5 – (66 ÷ 22) + 11 × 4 – 2 = 160

50 × 3 + 12 × 5 – 3 + 11 × 4 – 2 = 160

150 + 60 - 3 + 44 - 2 = 160

210 + 44 - 3 - 2 = 160

254 - 5 = 160

​It is incorrect.​

Option (d):12 and 11

50 × 3 + 11 × 5 – (66 ÷ 2) + 12 × 4 – 22 = 160

50 × 3 + 11 × 5 – 33 + 12 × 4 – 22 = 160

150 + 55 - 33 + 48 - 22 = 160

205 + 48 - 33 - 22 = 160 

253 - 33 - 22 = 160 

​It is incorrect.

Thus, the correct option is (a) 3 and 2.​