Solve the equation: 12Cx1+Cx+42=162, x∈N
12Cx1+Cx+42=162
i.e. 12×x!1!(x−1)!+(x+4)!2!(x+4−2)!=162
i.e. 12x(x−1)!(x−1)!+(x+4)!2(x+2)!=162
i.e. 12x+(x+4)(x+3)(x+2)!2(x+2)!=162
i.e. 12x+(x+4)(x+3)2=162
i.e. 24x+x2+7x+12=324
i.e. x2+31x−312=0
i.e. (x+39)(x−8)=0
⟹ x=−39 or x=8
∵ x \in N \;\; [given]
\implies x = 8 \; is the required solution.