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Question
6 boys and 6 girls sit in a row at random. Find the probability that all the girls sit together.
Solution
\[\text{ Here, 6boys and 6 girls can be arranged in a line in 12! ways } .\]
\[\text{ Total possible outcomes } = 12!\]
\[\text{ Consider 6 girls as a single element X } .\]
\[\text{ Now, 6 boys and X can be arranged in a line in 7! ways and girls can be arranged in 6! ways among them } .\]
\[P\left( \text{ all girls are together } \right) = \frac{7! \times 6!}{12!}\]
\[ = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\]
\[ = \frac{1}{11 \times 12}\]
\[ = \frac{1}{132}\]
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