There are four people in a room.

For every two people, there is a $50 \%$ chance that they are friends.

Two people are connected if:

- they are friends, or
- a third person is friends with both of them, or
- they have different friends who are friends of each other.

What is the probability that every pair of people in this room is connected?
(A) $\frac{18}{32}$
(B) $\frac{20}{32}$
(C) $\frac{22}{32}$
(D) $\frac{19}{32}$
(E) $\frac{21}{32}$
We label the four people in the room $A, B, C$, and $D$. We represent each person by a point. There are six possible pairs of friends: $A B, A C, A D, B C, B D$, and $C D$. We represent a friendship by joining the corresponding pair of points and a non-friendship by not joining the pair of points.

Since each pair of points is either joined or not joined and there are 6 possible pairs, then there are $2 \times 2 \times 2 \times 2 \times 2 \times 2=2^{6}=64$ possible configurations of line segments in the diagram. For example, if $A$ and $B, B$ and $C, A$ and $C$, and $B$ and $D$ are friends, then we obtain the configuration to the right.

Since each pair of points has equal probability of being joined or not joined, then each of the 64 possible configurations is equally likely, so has probability $\frac{1}{64}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-122.jpg?height=214&width=217&top_left_y=1335&top_left_x=1537)

Points $A$ and $B$, for example, are "connected" according to the definition, if they are joined, or if they are both joined to $C$, or if they are both joined to $D$, or if one is joined to $C$, the other to $D$ and $C$ and $D$ are joined. In other words, points $A$ and $B$ are connected if we can pass from $A$ to $B$ along line segments, possibly through one or both of $C$ and $D$.

If each pair of points is connected, we call the configuration fully connected.

Thus, we need to count the number of configurations that are fully connected.

First, we count the number of configurations including each possible number of line segments ( 0 through 6). After this, we will analyze each case to see which configurations are fully connected.

- 0 line segments: There is 1 such configuration.
- 6 line segments: There is 1 such configuration.
- 1 line segment: There are 6 such configurations, because we can choose any one of the 6 possible segments.
- 5 line segments: There are 6 such configurations, because we can start with all 6 line segments and remove any one of the 6 possible segments.
- 2 line segments: There are 15 such configurations. This is because there are 6 possible choices for the first line segment to add, and then 5 possible choices for the second line segment. This looks like $6 \times 5$ configurations, but each configuration is counted twice here (as we could choose $A B$ and then $B D$ or $B D$ and then $A B$ ). Therefore, there are $\frac{1}{2} \times 6 \times 5=15$ configurations.
- 4 line segments: There are 15 such configurations, because we can start with all 6 line segments and remove any of the 15 possible pairs of 2 line segments.
- 3 line segments: There are 20 configurations, because there are 64 configurations in total and we have already accounted for $1+1+6+6+15+15=44$ configurations.

Here is an example of each of these types of configurations:

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-123.jpg?height=298&width=1678&top_left_y=510&top_left_x=275)

6 segments
5 segments
4 segments
3 segments
2 segments
1 segment 0 segments

Now we analyze each of the possible classes of configurations to see if the relevant configurations are fully connected or not:

- 0 line segments

This configuration is not fully connected.

-6 line segments

This configuration is fully connected.

- 1 line segment

Each of these 6 configurations is not fully connected, since it can only have 2 points joined.

-5 line segments

Each of these 6 configurations is fully connected, since only one pair of points is not joined. If this pair is $A B$, for instance, then $A$ and $B$ are each joined to $C$, so the configuration is fully connected. (See the diagram labelled "5 segments" above.) The same is true no matter which pair is not joined.

-2 line segments

Each of these 15 configurations is not fully connected, since the two line segments can only include 3 points with the fourth point not connected to any other point (for example, $B$ connected to each of $A$ and $C$ ) or two pairs of connected points (as in the diagram labelled "2 segments" above).

-4 line segments

Each of these 15 configurations is fully connected.

Consider starting with all 6 pairs of points joined, and then remove two line segments.

There are two possibilities: either the two line segments share an endpoint, or they do not.

An example of the first possibility is removing $A B$ and $B C$ to get Figure 1. This configuration is fully connected, and is representative of this subcase.

An example of the second possibility is removing $A B$ and $C D$ to get Figure 2. This configuration is fully connected, and is representative of this subcase.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-123.jpg?height=220&width=219&top_left_y=2335&top_left_x=758)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-123.jpg?height=220&width=220&top_left_y=2335&top_left_x=1343)

Figure 2

Therefore, all 15 of these configurations are fully connected.

-3 line segments

There are 20 such configurations.

Each of these configurations involve joining more than 2 points, since 2 points can only be joined by 1 line segment.

There are several possibilities:

* Some of these 20 configurations involve only 3 points. Since there are three line segments that join 3 points, these configurations look like a triangle and these are not fully connected. There are 4 such configurations, which we can see by choosing 1 of 4 points to not include.
* Some of the remaining 16 configurations involve connecting 1 point to each of the other 3 points. Without loss of generality, suppose that $A$ is connected to each of the other points.

This configuration is fully connected since $B$ and $C$ are connected through $A$, as are $B$ and $D, C$ and $D$, and is representative of this

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-124.jpg?height=204&width=200&top_left_y=749&top_left_x=1708)
subcase.

There are 4 such configurations, which we can see by choosing 1 of 4 points to be connected to each of the other points.

* Consider the remaining $20-4-4=12$ configurations. Each of these involves all 4 points and cannot have 1 point connected to the other 3 points.

Without loss of generality, we start by joining $A B$.

If $C D$ is joined, then one of $A C, A D, B C$, and $B D$ is joined, which means that each of $A$ and $B$ is connected to each of $C$ and $D$. This type of configuration is fully connected.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-124.jpg?height=214&width=203&top_left_y=1308&top_left_x=1709)

If $C D$ is not joined, then two of $A C, A D, B C$, and $B D$ are joined. We cannot have $A C$ and $A D$ both joined, or $B C$ and $B D$ both joined, since we cannot have 3 segments coming from the same point.

Also, we cannot have $A C$ and $B C$ both joined, or $A D$ and $B D$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-124.jpg?height=212&width=217&top_left_y=1607&top_left_x=1694)
both joined, since we cannot include just 3 points.

Therefore, we must have $A C$ and $B D$ joined, or $A D$ and $B C$ joined. This type of configuration is fully connected.

Therefore, of the 64 possible configurations, $1+6+15+4+12=38$ are fully connected. Therefore, the probability that every pair of people in this room is connected is $\frac{38}{64}=\frac{19}{32}$.

ANSWER: (D)