Question

$\frac{{}^{8}C_0}{6}-{}^{8}C_1+{}^{8}C_2\times 6-{}^{8}C_3\times 6^2+...{}^{8}C_8\times 6^7$ is equal to

• A. $0$
• B. $6^7$
• C. $6^8$
• D. $\frac{5^8}{6}$

Answer 1

The correct option is D

$\frac{5^8}{6}$

Finding the value:

Given

$\frac{{}^{8}C_0}{6}-{}^{8}C_1+{}^{8}C_2\times 6-{}^{8}C_3\times 6^2+...{}^{8}C_8\times 6^7$

$=\frac{1}{6}[{}^{8}c_0-{}^{8}c_1\times 6$$+{}^{8}c_2\times 6^2-{}^{8}c_3\times 6^3+.....+{}^{8}c_3\times 6^8$

$=\frac{1}{6}{(1-6)}^8$

$=$$\frac{5^8}{6}$

Hence, Option(D) is correct.