<p>The coefficient of \(x^{10}\) in the expansion of \((1+x)^2(1+x^2)^3(1+x^3)^4\) is equal to</p>
Step-by-Step Solution
Key Concept: Expand each factor systematically and track which combinations of terms multiply to give x^10. The constraint is finding all non-negative integer solutions to 1·a + 2·b + 3·c = 10, where a ≤ 2, b ≤ 3, c ≤ 4.
<p><strong>Step 1:</strong> Write the expansion as selecting powers: $(1+x)^2(1+x^2)^3(1+x^3)^4 = \sum x^{a+2b+3c}$ where $\binom{2}{a}\binom{3}{b}\binom{4}{c}$ is the coefficient, with $0 \leq a \leq 2$, $0 \leq b \leq 3$, $0 \leq c \leq 4$.</p><p><strong>Step 2:</strong> Find all solutions to $a + 2b + 3c = 10$:</p><p>• $c=0$: $a+2b=10$ → $(a,b)=(0,5)$ invalid ($b>3$), $(2,4)$ invalid</p><p>• $c=1$: $a+2b=7$ → $(a,b)=(1,3)$ ✓, $(3,2)$ invalid ($a>2$)</p><p>• $c=2$: $a+2b=4$ → $(a,b)=(0,2)$ ✓, $(2,1)$ ✓, $(4,0)$ invalid</p><p>• $c=3$: $a+2b=1$ → $(a,b)=(1,0)$ ✓</p><p>• $c=4$: $a+2b=-2$ → No valid solutions</p><p><strong>Step 3:</strong> Sum coefficients: $\binom{2}{1}\binom{3}{3}\binom{4}{1} + \binom{2}{0}\binom{3}{2}\binom{4}{2} + \binom{2}{2}\binom{3}{1}\binom{4}{2} + \binom{2}{1}\binom{3}{0}\binom{4}{3}$</p><p>$= 2·1·4 + 1·3·6 + 1·3·6 + 2·1·4 = 8+18+18+8 = 52$</p><p>∴ Answer: 52 (Option A)</p>
Correct Answer: A