Mr. & Mrs. Anbouba recently attended a party, at which, there were three other couples. Various handshakes took place. No one shook hands with their spouse. No one shook his, or her, own hand. No one shook hands with the same person twice. At the end of the evening, Mrs. Anbouba asked the seven-other people, how many hands he, or she, had shaken. Each person gave her a different answer. How many hands did Mr. Anbouba shake?

### Identification of the problem

The crux of this problem, is to alternate extremes, so that we can eliminate possibilities. One way we can solve this problem, is to use a few diagrams to assist us in obtaining a solution.

### Analysis

For the first part of the problem, we need to figure out the maximum number of handshakes that are going to be possible. We know that there are two people in a couple. The problem states that Mr. & Mrs. Anbouba recently attended a party, at which, there were three other couples. From this, we can conclude that there must be four couples total. Mr. & Mrs. Anbouba being one couple, and three other couples, whose names we don’t know. This proves that we have a total of eight people.

The problem also states that no one shook hands with their spouse, and no one shook hands with themselves. Now that we have this information, we can form a simple math equation, that will tell us the maximum number of handshakes we will have.

We’ll take the total number of people at the party, and remove the two people (yourself and your spouse) that you cannot shake hands with. Now we have a formula we can use to solve the first part of the problem:

**Formula:** total people – (self + spouse) = max handshakes, **Or** 8 – (1 + 1) = 6

That formula tells us that the maximum number of handshakes, any one person can have, will be six. Now, we’ll need to figure out the minimum number of handshakes a person could have.

If you read the problem closely, you’ll notice that there is no constraint that requires every person to shake someone else’s hand. From this information, we can deduce that it’s possible for someone to have zero handshakes. This leaves us with zero for the minimum number of handshakes. We can use the following diagram, to represent the information we already know.

For this problem, we can have two different types of responses, one possible, and one absolute. I’ve listed them below:

### Responses

**Possible:** 0, 1, 2, 3, 4, 5, or 6, handshakes

**Absolute:** All the possible responses must be used at least once.

We now have enough information to draw the initial diagram, that we will use to solve this problem. The circles will represent the people, and the double-sided arrows, will represent the couples. Colored lines will represent the handshakes.

We know that someone is going to have six handshakes, so, I placed a number six, inside one of the circles, as a starting point. It should be noted, that I chose this circle at random. I could have chosen any of the circles as a starting point, and the result would have been the same. I’ve also crossed-off the number six, in the list, to make it clear that the number has already been used.

The next thing we want to do, is draw a line between person number six, and all the people they will shake hands with. We know that person number six can’t shake hands with their spouse, so we won’t be drawing a line to that person. The next diagram shows the result:

We can tell that person number six, has six handshakes. The spouse of person number six, doesn’t have any handshakes, and everyone else has one handshake.

We can figure out the next part of the problem, by alternating extremes. In other words, we’ll try to figure out what person will have zero handshakes. By looking at the diagram above, we can tell that only one person can have zero handshakes. We know this, because person number six already has six handshakes, and everyone else already has one handshake. We can identify that person in the next diagram:

Now, someone needs to have five handshakes. It doesn’t matter at this point who we pick to have five handshakes. So, I’ll pick another person at random. Once I’ve identified that person, I’ll go ahead and draw the handshake lines, and mark the number five, as being used. You can see the result in the next diagram:

The blue lines indicate the four-additional people, that person number five has shaken hands with. Notice that they did not shake hands with their spouse, or the spouse of person number six, because we’ve already stated that person won’t have any handshakes.

We can now figure out who will have only one handshake. We can tell from the next diagram, that it’s only possible for the spouse of person number five to have one handshake, because everyone else already has more than one. This is represented in the next diagram.

For person number four, we can pick any of the unlabeled people, because they all have less than four handshakes. I’ll place person number four at the top, and draw the handshake lines, to the two people on the left, as those are the only two people, that person number four can shake hands with. Remember, that they can’t shake hands with their spouse, and everyone else is already labeled with their corresponding handshakes.

If we look at the diagram above, we can see that the unlabeled person (right) on the top, currently has two handshakes. So, we can mark the remaining person on the top, as being person number two. If you look at the two-unlabeled people on the left side, you’ll notice both have three handshakes. Let’s update the diagram with the changes for person number two, and take a closer look at the remaining two people on the left:

As stated above, both people on the left, currently have three handshakes, that are identified by a red, a blue, and a green, line. We can also tell by looking at our number list, that we only have one number left to use; the number three. Let’s label the person on the top-left, as person number three, and see what happens:

Looking at the diagram above, we can tell that there is only one unlabeled person left. We are also out of numbers to use. If we count the lines on the unlabeled person, we can tell they have three lines, one red, one blue, and one green. This means that this person also has three handshakes. We can now complete our diagram:

The final diagram shows us that we have two people at the party, that will have three handshakes. If we look back at the original question, it states that; at the end of the evening, Mrs. Anbouba asked the seven-other people how many hands he or she had shaken, and each person gave her a different answer.

When we look at the diagram, we can tell that Mrs. Anbouba must have been one of the people on the left. We know this, because if she was any other person, then two of the people at the party would have reported the same number of handshakes, when she asked at the end of the evening. Two of the people would have reported they shook hands three times.

The original problem also states that she asked the seven-other people at the party, how many times they shook hands. This tells us that she did not respond to the question. Otherwise there would have been eight responses to the question.

We don’t know what person on the left is Mrs. Anbouba, but it doesn’t matter. We only need to know who her spouse is, because the question is asked, how many times Mr. Anbouba, shook hands. Since we know that Mrs. Anbouba was one of the people on the left, we also know that, Mr. Anbouba must have been one of the people on the left, because, he was her spouse.

With this information, we can conclude that the total number of handshakes Mr. Anbouba received, was three.

### Heuristics

- We made several diagrams.
- We listed the possible responses.
- We calculated the minimum, and maximum responses.
- We alternated extremes.
- We eliminated possibilities.
- We created a formula to calculate the max handshakes.

### Looking Back

When looking at a problem like this, the solution can seem impossible. It can be hard to picture in your mind, what the outcome might be. To get around this, we first calculated the minimum, and maximum, responses, and listed what the possible response types were.

We talked through the problem, while using diagrams to help break the problem down, into more manageable pieces. When using these diagrams, we’re able to get a better understanding of what was going on at the party. We could conclude that Mrs. Anbouba, was, in fact, a duplicate.

We also used the method of alternating extremes, to make solving the problem easier. This helped us eliminate possibilities, while completing the diagram. If we would have worked from highest to lowest, (or lowest to highest) we would have ran into complications solving the problem. Try doing this problem again, without alternating extremes; you’ll see that it becomes a lot harder to eliminate the possibilities.

While it’s possible to solve this problem without the use of diagrams, it would have been a lot harder to come up with the correct answer, and there wouldn’t have been an easy way to test that the answer was correct.

There’s a terrific amount of kndeowlge in this article!