Coefficient of x11 in the expansion of (1+x2)4(1+x3)7(1+x4)12

2x1 + 3x2 + 4x3 = 11
Possibilities are (0,1,2); (1,3,0); (2,1,1); (4,1,0).
∴Required coefficients = (4C0 × 7C1 × 12C2) + (4C1 × 7C3 × 12C0) + (4C2 × 7C1 × 12C1) + (4C4 × 7C1 × 12C0)
= (1 × 7 × 66) + (4 × 35 × 1) + (6 × 7 × 12) + (1 × 7 × 1)
= 462 + 140 + 504 + 7
= 1113.