<p>Find the coefficient of \(a^3b^4c^5\) in the expansion of \((bc + ca + ab)^6\).</p>
Step-by-Step Solution
Key Concept: Recognize that (bc + ca + ab)^6 expands using the multinomial theorem, where you need to select terms that multiply to give exactly a³b⁴c⁵. Each selection from the 6 factors must contribute specific powers of a, b, and c.
<p><strong>Step 1:</strong> Use multinomial theorem. We need (bc)^x · (ca)^y · (ab)^z where x + y + z = 6.</p><p><strong>Step 2:</strong> This gives a^(y+z) · b^(x+z) · c^(x+y). For a³b⁴c⁵, we need:</p><ul><li>y + z = 3 (power of a)</li><li>x + z = 4 (power of b)</li><li>x + y = 5 (power of c)</li></ul><p><strong>Step 3:</strong> Solving this system: From equations, (y+z) + (x+z) + (x+y) = 3 + 4 + 5 = 12, so 2(x+y+z) = 12, giving x+y+z = 6 ✓</p><p>Solving: x = 2, y = 3, z = 1</p><p><strong>Step 4:</strong> The coefficient is the multinomial coefficient: $$\frac{6!}{2! \cdot 3! \cdot 1!} = \frac{720}{2 \cdot 6 \cdot 1} = \frac{720}{12} = 60$$</p><p>∴ Answer: <strong>60</strong></p>
Correct Answer: 60