1. The problem is to calculate $4\frac{5}{6} \times \left(2\frac{1}{4} + 1\frac{1}{3}\right)$.\n\n2. Convert mixed numbers to improper fractions:\n$4\frac{5}{6} = \frac{4 \times 6 + 5}{6} = \frac{24 + 5}{6} = \frac{29}{6}$\n$2\frac{1}{4} = \frac{2 \times 4 + 1}{4} = \frac{8 + 1}{4} = \frac{9}{4}$\n$1\frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{3 + 1}{3} = \frac{4}{3}$\n\n3. Add the fractions inside the parentheses:\n$\frac{9}{4} + \frac{4}{3} = \frac{9 \times 3}{4 \times 3} + \frac{4 \times 4}{3 \times 4} = \frac{27}{12} + \frac{16}{12} = \frac{43}{12}$\n\n4. Multiply the fractions:\n$\frac{29}{6} \times \frac{43}{12} = \frac{29 \times 43}{6 \times 12} = \frac{1247}{72}$\n\n5. Convert the improper fraction back to a mixed number:\nDivide 1247 by 72:\n$1247 \div 72 = 17$ remainder $31$\nSo, $\frac{1247}{72} = 17\frac{31}{72}$\n\nFinal answer: $17\frac{31}{72}$