Q. The number of like terms in
\(\frac{1}{4}\)a²bc, – \(\frac{2}{3}\)bca², \(\frac{2}{5}\) ba²c, – \(\frac{1}{2}\)cba² is
\(\frac{1}{4}\)a²bc, – \(\frac{2}{3}\)bca², \(\frac{2}{5}\) ba²c, – \(\frac{1}{2}\)cba² is
✅ Correct Answer: (A)
4