If \(^9C_x = 4[^7C_{x – 1}]\), find the values of \(x\)
Explanation
\(^9C_x = 4[^7C_{x - 1}]\)
\(\frac{9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(7 - (x - 1)!)}\)
\(\frac{9!}{x!(9 - x)!} = 4*\frac{7!}{(x - 1)!(8 - x)!}\)
\(=\frac{9 \times 8 \times 7!}{x(x - 1)!(9 - x)(8 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)
Cancel out the common terms
\(=\frac{9 \times 8}{x(9 - x)}=\frac{4}{1}\)
\(=4x(9-x) = 9*8\)
\(=36x- 4x^2=72\)
divide thru by 4
\(=x^2 - 9x+18 = 0\)
\(=x^2-6x-3x+18=0\)
\(=x(x-6)-3(x-6)=0\)
\(=(x-6)(x-3)=0\)
\(\therefore x\) = 6 or 3