- Home
- Standard 11
- Mathematics
14.Probability
medium
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
Solution
(b) $6$ 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}}$.
Standard 11
Mathematics
