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(B) Total ways to divide $75$ students into two sections of $30$ and $45$ is $^{75}C_{30}$.Let the two particular students be $S_1$ and $S_2$.Case $I$

Vedclass · Accepted Answer

$\frac{19}{37}$

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(B) Total ways to divide $75$ students into two sections of $30$ and $45$ is $^{75}C_{30}$.Let the two particular students be $S_1$ and $S_2$.Case $I$: Both $S_1$ and $S_2$ are in the section of $30$ students. The number of ways is $^{73}C_{28}$.Case $II$: Both $S_1$ and $S_2$ are in the section of $45$ students. The number of ways is $^{73}C_{43}$.The required probability is $\frac{^{73}C_{28} + ^{73}C_{43}}{^{75}C_{30}}$.Using the formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we get:$\frac{\frac{73!}{28!45!} + \frac{73!}{43!30!}}{\frac{75!}{30!45!}} = \frac{73!}{75!} 	imes \left( \frac{30!45!}{28!45!} + \frac{30!45!}{43!30!} ight)$$= \frac{1}{75 	imes 74} 	imes (30 	imes 29 + 45 	imes 44)$$= \frac{870 + 1980}{5550} = \frac{2850}{5550} = \frac{285}{555} = \frac{19}{37}$.

If two sections of strengths $30$ and $45$ are formed from $75$ students who are admitted in a school,then the probability that two particular students are always together in the same section is

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