Answer to Question #90713 in Algebra for Sayem

Question

2nCn * 3^-n = 6C3 * 2^3 * 6^-3
Then n=?

Answer 1

$_{2n}C_n=_6C_3$ so, n=3.

$3^{-n}=\frac{2^3}{6^3}=\frac{1}{3^3}=3^{-3}$ so, n=3.

Thus, the only solution is n=3.