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Algebra - Elements of Combinatorics

Solve the equation: 12Cx1+Cx+42=162, xN


12Cx1+Cx+42=162

i.e. 12×x!1!(x1)!+(x+4)!2!(x+42)!=162

i.e. 12x(x1)!(x1)!+(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+31x312=0

i.e. (x+39)(x8)=0

x=39 or x=8

x \in N \;\; [given]

\implies x = 8 \; is the required solution.