Question

$10$ distinct balls are arranged in a row. The number of ways of selecting three of these balls so that no two of them are next to each other is

• A. $\frac{1}{6}\times 8\times 7\times 6$
• B. ${}^8{C}_3$
• C. ${}^7{C}_3{+}^7{C}_2$
• D. None of these

Answer 1

The correct option is A $\frac{1}{6}\times 8\times 7\times 6$

Selecting $n$ objects such that no two objects are together, that can be done in ${}^{n-2}{C}_3$ ways

Here $n=10$

So, total no of w${=}^{10-2}{C}_3=\frac{8!}{3!5!}=\frac{8\times 7\times 6\times 5!}{3\times 2\times 5!}=\frac{1}{6}$$\times 8\times 7\times 6$