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InterviewSolution
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6 boys and 6 girls are seated in a row. Probability that all the boys sit together is(a) \(\frac{1}{102}\)(b) \(\frac{1}{112}\)(c) \(\frac{1}{122}\)(d) \(\frac{1}{132}\) |
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Answer» (d) \(\frac{1}{132}\) Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the remaining, i.e., 6 girls can be arranged in a row in 6! ways. There are 5 places between the 6 girls and 2 on the extreme ends, where the entity of 6 boys can be placed, i.e., for the single entity of 6 boys, we have 7 places where they can be arranged in 7 ways and also amongst themselves they can be arranged in 6! ways. ∴ No. of ways of arranging 6 boys and 6 girls in a row where the 6 boys are together = 6! × 7 × 6! ∴ Required probability = \(\frac{6!\times7\times6!}{12!}\) = \(\frac{7\times6\times5\times4\times3\times2}{12\times11\times10\times9\times8\times7}\) = \(\frac{1}{132}\). |
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