Six boys and six girls sit in a row at random.

Question

Professor PC Thomas Classes · Answer

(i) 6 boys and 6 girls, total 12 items can be arranged in a row in ${ }^{12} P_{12}=\lfloor 12$ ways. Put the 6 girls in a bundle. We have now 7 things, i.e., 6 boys and one bundle. The seven objects can be arranged in a row in ${ }^{7} P_{7}=\boxed{7}$ ways the 6 girls in the bundle can be arranged in ${ }^{6} P_{6}$ ways.\ $	herefore$ Number of arrangements with the 6 girls togeiner $=\underline{6} 	imes \underline{\angle}$.\ Hence the r., ired probability\ $=\frac{\underline{\underline{6} 	imes \underline{7}}}{\underline{12}}=\frac{\underline{\underline{6}}}{8 	imes 9 	imes 10 	imes 11 	imes 12}=\frac{1}{132}$\ (ii) Now the boys and girls are to sit alternately. If we begin with a boy, there will be 6 places for the 6 girls and the six girls can be arranged in the six places in 6 ways while the 6 boys can be arranged in the 6 places in 6 ways.\ $	herefore$ Number of ways $=\underline{6} 	imes \underline{6}$.\ Similarly if we begin with a girl, they can sit alternately in $\underline{6} 	imes \underline{6}$ ways.\ $	herefore \quad$ Requiredprobability $=\frac{2 	imes \underline{6} 	imes \underline{6}}{\underline{12}}=\frac{1}{462}$\