Since, $\left(1+x\right)^{n }= ^{n}C_{0}+^{n}C_{1}x+^{n}C_{2} x^{2} +...+^{n}C_{n} x^{n}$
Put x = 1 and n = 10, we get
$2^{10} = ^{10}C_{0}+^{10}C_{1}+^{10}C_{2}+...+^{10}C_{10}$
$2^{10} = 1+^{10}C_{1}+^{10}C_{2} +...+^{10}C_{9}+1$
$2^{10} - 2 = ^{10}C_{1}+^{10}C_{2} +...+^{10}C_{9}$