A building has a ground floor and 10 more flo | vedclass

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(C) The building has $10$ floors above the ground floor, but the lift does not stop at the $1$st and $2$nd floors. Thus, there are $10 - 2 = 8$ floors

Vedclass · Accepted Answer

$7056$

Vedclass · Answer

(C) The building has $10$ floors above the ground floor, but the lift does not stop at the $1$st and $2$nd floors. Thus, there are $10 - 2 = 8$ floors available for exiting.Nine persons enter the lift. We need to select $4$ persons to exit at one floor and the remaining $5$ persons to exit at a different floor.The number of ways to partition $9$ persons into a group of $4$ and $5$ is $\binom{9}{4} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.The number of ways to choose $2$ distinct floors from the $8$ available floors is $P(8, 2) = 8 	imes 7 = 56$ (since the order of floors matters, i.e.,which group exits at which floor).The total number of ways is $126 	imes 56 = 7056$.

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