# Why did we multiply (b) by 5! and 6!?

Six people enter an elevator at the ground floor of a 12-storey building. They choose their exit floors independently of each other.

Why did we multiply (b) by 5! and 6!? Does the order matters? If so, why didn't we take into consideration the order in (a).

• In both parts, it does matter who gets off the elevator on which floor. – N. F. Taussig Jul 23 '18 at 14:22
• So why then, we did not multiply by 5! And 6! In a)? – Rayri Jul 23 '18 at 14:25

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Find the probability that exactly two people get out on one floor, exactly two people on another floor, and each of the remaining people gets off where no one else does.

Since each of the six people can choose to exit the elevator on one of the twelve floors, there are $12^6$ elements in the sample space.

There are $\binom{12}{2}$ ways of selecting the two floors at which two people each exit the elevator and $\binom{10}{2}$ ways of selecting at which two of the remaining ten floors one person each exits the elevator. There are $\binom{6}{2}$ ways of selecting which two people exit the elevator on the lower of the two floors from which two people exit the elevator and $\binom{4}{2}$ ways of selecting which two of the remaining four people exit the elevator on the higher of those floors. There are $\binom{2}{1}$ ways of selecting which one of the remaining two people exits the elevator on the lower of the two floors at which one person exits the elevator. The remaining person must exit the elevator on the higher of the two floors at which one person exits the elevator. Hence, the number of favorable cases is $$\binom{12}{2}\binom{10}{2}\binom{6}{2}\binom{4}{2}\binom{2}{1}$$

Notice that it matters which person exits the elevator on which floor.

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Assume instead each person exits the top floor with probability $1/2$ and with equal probability at each of the other floors. Determine the probability that they all get out at different floors.

It does matter who exits the elevator on which floor.

Since the probability that a person exits the elevator on the top floor is $1/2$, the probability that she or he does not exit the elevator on the top floor is $1 - 1/2 = 1/2$. Since it is equally likely that the person exits the elevator on any of those eleven floors, the probability that a person exits at a particular lower floor is $$\frac{1}{11} \cdot \frac{1}{2} = \frac{1}{22}$$

There are two possibilities for the favorable cases. Either nobody gets off the elevator on the top floor or one person does.

We find the probabilities of each case.

Case 1: Nobody exits at the top floor.

There are $\binom{11}{6}$ ways to select at which six of the eleven lower floors the six people exit the elevator and $6!$ ways to assign them to those floors. Each of those six people has probability of $1/22$ of exiting the elevator at that particular floor. Since these choices are independent, the probability that six people exit the elevator at six different floors, none of which is the top floor, is $$\binom{11}{6}6!\left(\frac{1}{22}\right)^6$$

Case 2: One person exits at the top floor.

There are six ways to select the person who exits the elevator at the top floor. The probability that he or she exits the elevator on that floor is $1/2$. Each of the other people must exit the elevator at a different one of the lower eleven floors. There are $\binom{11}{5}$ ways to select the five floors at which they exit and $5!$ ways to arrange them on those floors. Each of those five people has probability $1/22$ of selecting that particular floor. Since these choices are independent, the probability that each person exits the elevator at a different floor if one of them chooses to exit at the top floor is $$\binom{6}{1}\left(\frac{1}{2}\right)\binom{11}{5}\left(\frac{1}{22}\right)^5$$

Since these cases are mutually exclusive, the desired probability is obtained by adding the results for the two cases.

The solution may have been more transparent if instead of writing what they did, they wrote

$$_{11}P_6\left(\frac{1}{22}\right)^6 + \left(6 \times \frac12\right){} \left[ _{11}P_5 \left(\frac{1}{22}\right)^5\right]$$

for that is equal to the expression in (b). From left to write, we have

• Case I: No one gets off on the top floor
• $_{11}P_6$ is the number of permutations of $6$ out of the $11$ non-top floors
• $1/22$ is the probability of getting out at any non-top floor
• Case II: One person gets off on the top floor
• $6$ is the number of ways to select the person who gets out on the top floor
• $1/2$ is the probability of getting out on the top floor
• $_{11}P_5$ is the number of permutations of $5$ out of the $11$ non-top floors
• $1/22$ is again the probability of getting out on any non-top floor

Note that

$$_nP_k = k!_nC_k = k!\binom{n}{k}$$

We take permutations—in other words, the order matters here—because assigning (for instance) floor $5$ to person $1$ and floor $7$ to person $2$, etc., is different from assigning floor $7$ to person $1$ and floor $5$ to person $2$, etc.

I'm still trying to parse the answer to (a). I have to say, I'm afraid, their presentation is not the best. ETA: OK, I've now got it, but I see that saulspatz has anticipated me and explained part (a) nicely.

In part a) they're still taking into account who gets off at which floor. There are ${12 \choose 2}$ ways to pick the two floors at which two people get off and ${10\choose2}$ ways to choose the floors at which single people get off. Then there are ${6\choose2}$ ways to assign two people to the first "double" floor, ${4\choose2}$ ways to assign two people to the second double floor, and ${2\choose1}$ ways to assign two people to the "single" floors.