The elevator starts with seven passengers and stops at ten floors. The various arrangements of discharge may be denoted by symbols like $(3,2,2)$, to be interpreted as the event that three passengers leave together at a certain floor. two other passengers at another floor, and the last two at still another floor. Find the probabilities of the fifteen possible arrangements ranging from $(7)$ to $(1, 1, 1, 1, 1, 1, 1)$.
The answer, for example for $P(5,2)= \frac{10!} {8!*1!*1!} *\frac{7!} {5!*2!}*10^{-7}$
Please, can you explain the first term. I mean why $10$ floors are divided into groups with $8, 1$ and $1$ floors. Why not $9,1$ or anything else?
Best Answer
It's because the $(5,2)$ discharge means that all passengers get off on one of two floors. So, that could be $5$ in floor $1$, and $2$ on floor $2$, or $5$ on floor $1$ and $2$ on floor $3$, or $5$ on floor $2$, and $2$ on floor $1$ ... etc.
There are $\frac{10!} {8!*1!*1!}$ ways to pick those two floors: $8$ floors where no one gets off, $1$ floor where $5$ people get off, and $1$ floor where $2$ people get off.