(1+x)2(1−x)3=(1+x)2 (1−x)−3
=(1+x2+2x)(1−x)−3
We know that coefficient of xr in (1−x)−n= n+r−1Cr
∴x4 coefficient
=1×x4 coefficient in (1−x)−3+2×x3 coefficient in (1−x)−3+1×x2 coefficient in (1−x)−3
=1×3+4−1C4+2×3+3−1C3+1×3+2−1C2
=6C4+2×5C3+4C2
=15+20+6
=41
Alternate Solution:
(1+x)2(1−x)3=(1+x)2 (1−x)−3=(1+2x+x2)(1+3x+6x2+10x3+15x4+⋯)
Hence x4 coefficient will be
=1⋅15+2⋅10+1⋅6=41