Not for the weak-hearted

The following 4 possibilities can occur :

$(x^2)^0 \times (x^3)^1 \times (x^4)^2$

$(x^2)^1 \times (x^3)^3 \times (x^4)^0$

$(x^2)^2 \times (x^3)^1 \times (x^4)^1$

$(x^2)^4 \times (x^3)^1 \times (x^4)^0$

Since there are $4$ terms of $(1+x^2)$ , $7$ termsof $(1+x^3)$ , $12$ terms of $(1+x^4)$ . sum of coefficients would be - $\binom{4}{0} \binom{7}{1} \binom{12}{2} + \binom{4}{1} \binom{7}{3} \binom{12}{0} + \binom{4}{2} \binom{7}{1} \binom{12}{1} + \binom{4}{4} \binom{7}{1} \binom{12}{0}$ $462 + 140 + 504 + 7 = \boxed{1113}$

• 11 = 3+2+2+2+2
• =3+2+2+4
• =3+4+4
• =3+3+3+2 So the coefficient of $x^{11}$ is [4C4] [7C1] [12C0]+[4C2] [7C1] [12C1]+[4C0] [7C1] [12C2]+[4C1] [7C3] [12C0]=1113