Find out the two signs to be interchanged for making following equation correct.\(\frac{1}{4} \div \frac{1}{{64}}\; - 5 + 45 \times 120 = 5\)

1) + and ×\(\frac{1}{4} \div \frac{1}{{64}}\; - 5 + 45 \TIMES 120 = 5\)After interchanging signs:\(\frac{1}{4} \div \frac{1}{{64}}\; - 5\; \times 45 + 120\) ≠ 52) × and –\(\frac{1}{4} \div \frac{1}{{64}}\; - 5 + 45 \times 120 = 5\)After interchanging signs:\(\frac{1}{4} \div \frac{1}{{64}}\; \times 5 + 45 - 120 = 5\)3) + and –\(\frac{1}{4} \div \frac{1}{{64}}\; - 5 + 45 \times 120 = 5\)After interchanging signs:\(\frac{1}{4} \div \frac{1}{{64}} + 5 - 45 \times 120\; \ne {\rm{\;}}5\)4) – and ÷\(\frac{1}{4} \div \frac{1}{{64}}\; - 5 + 45 \times 120 = 5\)After interchanging signs:\(\frac{1}{4} - \frac{1}{{64}}\; \div 5 + 45 \times 120 \ne {\rm{\;}}5\)HENCE, the correct answer is ‘× and –‘.