Question

$6$ boys and $4$ girls sit around a table at random. The probability that no two girls sit together is

• A. $\frac{1}{126}$
• B. $\frac{4}{14}$
• C. $\frac{5}{42}$
• D. $\frac{1}{21}$

Answer 1

Answer: (C)

$\frac{5}{42}$

From given, we have,

• Considering boys sit first, then the first girl has to pick a seat between two boys.
• The second girl could choose any of the seven seated people to be on her right, five choices of which mean she is not sitting next to the first girl.
• Assuming she makes one of those choices, the third girl can choose from eight people, four of which mean she is between two boys.
• The fourth girl can choose from nine seats, three of which mean she is between two boys.

$\frac{5}{7}\times \frac{4}{8}\times \frac{3}{9}=\frac{60}{504}=\frac{5}{42}$