## Task B-4.6.

An elevator can transport a maximum of eight people in one ride. Eleven residents, including two who are feuding, are waiting for the elevator, and they are all going to the same floor. What is the probability that the feuding residents will be in the elevator if all residents must be transported by the elevator in the minimum number of rides?
## Solution.

Eleven residents can be transported in at least two trips.

First, let's determine the total number of ways to transport all residents, regardless of the quarreling pair.

In two trips, the elevator can transport:

3 residents in the first trip + 8 residents in the second trip or vice versa $8+3$, then $4+7$ or $7+4$, and $5+6$ or $6+5$. In the first case, out of 11 residents, we choose 3 who go in the first trip (or 8 who go in the second trip) in $\binom{11}{3}=\binom{11}{8}=165$ ways. Clearly, the number of ways is the same if 8 residents go in the first trip and 3 in the second, so the total number of ways for the transport of $3+8$ and $8+3$ is $2 \cdot \binom{11}{3}=330$.

The number of ways for the elevator to transport 4 residents in the first trip and 7 in the second trip or vice versa is $2 \cdot \binom{11}{4}=660$.

The number of ways for the elevator to transport 5 residents in the first trip and 6 in the second trip or vice versa is $2 \cdot \binom{11}{5}=2 \cdot 462=924$.

Thus, the total number of ways is $2\left(\binom{11}{3}+\binom{11}{4}+\binom{11}{5}\right)=1914$.

1 point

Next, let's determine the number of ways in which the quarreling pair of residents are together in the elevator.

In each of the trip combinations $(3+8,8+3,4+7,7+4,5+6,6+5)$, the quarreling residents go in the elevator together either in the first or the second trip, and it is sufficient to choose the remaining residents in the elevator from the remaining 9 residents.

If the elevator in the first trip transports 3 residents, including the quarreling pair, we need to choose only one from the remaining 9 residents, which can be done in $\binom{9}{1}=9$ ways. If the elevator in the first trip transports 3 residents, but not including the quarreling pair, the number of ways is equal to the number of ways to choose 3 residents from the remaining 9, i.e., $\binom{9}{3}=84$.

If the elevator in the first trip transports 8 residents, including the quarreling pair, we choose 6 residents from the remaining 9 in $\binom{9}{6}=\binom{9}{3}=84$ ways. If the quarreling pair is in the second trip, the number of ways to choose the remaining resident is $\binom{9}{1}=9$.

2 points

The total number of ways in the trip combinations $3+8$ and $8+3$ is $2\left(\binom{9}{1}+\binom{9}{6}\right)=186$.

We will similarly calculate for the remaining cases.

In the trip combinations $4+7$ and $7+4$, it is $2\left(\binom{9}{2}+\binom{9}{5}\right)=324$.

In the trip combinations $5+6$ and $6+5$, it is $2\left(\binom{9}{3}+\binom{9}{4}\right)=420$.

2 points

Finally, the probability that the quarreling residents will be together in the elevator is:

$$
\frac{\left(\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+\binom{9}{4}+\binom{9}{5}+\binom{9}{6}\right) \cdot 2}{1914}=\frac{930}{1914}=\frac{155}{319}
$$

1 point