Question

Six boys and six girls sit in a row randomly. The probability that the six girls sit together

• A. $\frac { 1 } { 77 }$
• B. $\frac { 1 } { 132 }$
• C. $\frac { 1 } { 231 }$
• D. None of these

Answer 1

Answer: (B)

$\frac { 1 } { 132 }$

Answer 2

6 boys and 6 girls can be arranged in a row in $12 !$ ways. If all the 6 girls are together, then the number of arrangement are $7 ! \times 6 !$ .

Hence required probability$= \frac { 7 ! .6 ! } { 12 ! }$

$= \frac { 6 \times 5 \times 4 \times 3 \times 2 } { 12 \times 11 \times 10 \times 9 \times 8 } = \frac { 1 } { 132 }$.