If \(^9C_x = 4[^7C_{x - 1}]\), find the values of \(x\)

Explanation

\(^9C_x = 4[^7C_{x - 1}]\)

\(\frac{9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(7 - (x - 1)!)}\)

\(={9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)

\(=\frac{9 \times 8 \times 7!}{x(x - 1)!(9 - x)(8 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)

Cancel out the common terms

\(=\frac{9 \times 8}{x(9 - x)}=\frac{4}{1}\)

\(=x(9-x)=9*8\)

\(=9x-x^2=72\)

\(=x^2-9x+72=0\)

\(=x^2-6x-3x+18=0\)

\(=x(x-6)-3(x-6)=0\)

\(=(x-6)(x-3)=0\)

\(\therefore x\) = 6 or 3