6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th
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(d) \(rac{1}{132}\)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 = \(rac{6! imes7 imes6!}{12!}\) = \(rac{7 imes6 imes5 imes4 imes3 imes2}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{132}\).
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