Question

Six boys and six girls are to sit in a row at random. The probability that boys and girls sit alternately is

• A. $\frac{6!{\times }^7{p}_6}{12!}$
• B. $\frac{6!6!}{12!}$
• C. $\frac{2\times 6!6!}{12!}$
• D. $\frac{2\times 6!}{12!}$

Answer 1

The correct option is C $\frac{2\times 6!6!}{12!}$

The two ways of doing this : $BGBGBGBGBGBG$ or $GBGBGBGBGBGB$
Since all boys and girls are distinct.
The total ways of arrangements $=12!$
The required probability $=2\ast 6!\ast 6!/(12!)$