A building has ground floor and 10 more floors.

Question

A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the $10^{	ext {th }}$ floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit ...

Accepted Answer

First, decide which $4$ out of the $9$ persons will get down together. The remaining $5$ persons are then fixed automatically.

Number of ways to form the two groups of sizes $4$ and $5$ is $\frac{9!}{4!5!}=126$

Now, the lift does not stop at the ${1}^{}$ and ${2}^{}$ floors. So people can get down only on floors $3$ to $10$, i.e., $8$ possible floors.

Choose any $2$ different floors out of these $8$ floors for the two groups to exit: ${}^{8}{C}_{2}=28$

After choosing the two floors, we must decide which floor is for the group of $4$ and which floor is for the group of $5$. This can be done in $2$ ways.

So, total number of ways is $126	imes 28	imes 2=7056$

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