Fancy Parties

For every friendship there have to be two people, one on each end of the relationship, who are being friends.

So the sum of all the "friends" of all the people has to be even.

$3\times3+2\times2=13$

Since 13 is odd, it may not be an answer, so such situation is impossible.

If a person has n friends, then there must be n edges attached to that person in the corresponding friendship graph (in other words, the degree of the node is n). The total degree of the graph is the sum of the degrees of the individual nodes.

Adding an edge to the graph contributes +2 to the total degree (+1 to the degrees of the two nodes that the edge connects). Therefore the total degree must be even, since it is equal to 2 times the number of edges.

The situation described corresponds to 3 nodes of degree 3, and 2 nodes of degree 2, which gives: $3 \times 3 + 2 \times 2 = 13 \neq 2\times\text{number of edges}$ which is odd, and therefore impossible.

We can visualize the problem by creating a case (you may choose anyone and the eventual outcome will be the same) for the following 5 persons:

A B C D E

1. Let A have 3 friends: B C D

2. Let B have 3 friends: A C D (A was already connected to B in 1.)

3. Let C have 3 friends: A B E (C was already connected to A and B in 1. and 2.)

We have fulfilled the 3 persons 3 friends criteria.

1. This means D definitely has 2 friends: A and B as per the above conditions.

2. Similarly, E has C as a friend. Now E cannot add either A, B or D as a friend because it will violate the 3 persons 3 friends condition in order to generate 2 persons 2 friends condition.

Thus, it is impossible to achieve such an overall condition.

In the language of Graph Theory,

Sum of degrees of the vertices of an undirected graph is always even.

Now, consider the people in the party as the vertices of a graph with number of friends of a person as the degree of the vertex representing the person.

Then $3$ people with $3$ friends and $2$ people with $2$ friends leave us with $3 \times 3 + 2 \times 2 = 13$ (which is an odd number) as total degrees of vertices.

So, $\boxed{\text{NO}}$ is the answer.

Start with Five People Pick any person to have three friends. Now we have choices. We can make our second three friend guy a friend or not. Let's make him not a friend. Now we have three friends who have two friends, and two with three. Unfortunately, whoever we choose to upgrade to three friends will then have another friend with three friends, so it won't work.

So we try making him a friend of our first guy, He can only be friends with first guy's friends, or with the other one. Let's have him with first guy's friends only

Then, our third friend with three firends is a friend of the first firend, or not. If he's not, then we get too many friends. If he is, then we have a lonely friend :(

If our second friend is friends with the last friend, then we are again forced to pick the third three-friend to have too many friends.

Sum of all the "friends" has to be even. 3 3 + 2 2 = 13 (odd) So it is not Possible, so such situation is impossible.

according to the graph,

$\text{A \& E is directly connected with B,C,D-----3 people}$

$\text{B,C, \& D is directly connected with A,E--------2 people}$

$\text{it is clearly seen that,2 people with 3 friends is possible but 3 people with 3 friends is not possible in 2 dimension}$

not possible, simply because there are just 5 people!

we have only five people so for 3friends and 3 people it will be 6people

There are a theorem in graphs theory, and says; in a graph, there is an even number of vertex of odd degree. So, in the problem there are 3 people with 3 friends, in a graph, thus is, an odd number of vertex of odd degree.

Assuming that 3 unique persons are going to befriend 3 similar others, this means there should be at least 6 persons, hence it is impossible.

I DO NOT WANT TO COMMENT ON THIS SOLUTION (the odd number of 3x3 + 2x2 which I hadn't noticed is a very good solution!). I WISH TO ASK A QUESTION! Which is how do you make a comment about a question before you have answered it. In particular what does the notation of a line over 3 variables abc mean which is the notation used in the question of this week. (I suppose that I'll just come back here to see if anyone has been kind enough to answer my question?) The general question that I've asked is a valid one I think. Is there a 'platform' to ask questions such as 'what does this notation mean'? Regards, David

There are 5 persons in the group (including yourself). The maximum number of friends you could have is 4 (you have to exclude "yourself"). Thus n-1. 5 -1 = 4. Given the supplied network diagram, it would be impossible.

Seeing all these mathematical formulas...it's way simpler:

If three people have three friends the total would be 6 people. Can't be cause there are 5 people total...

The problem can be modelled as an undirected graph. In essence one uses the Handshake-Theorem, which says that the sum of all the numbers of neighbors any vertex has (called degree of the vertex) must be two times the number of total edges in the graph. This implies that an undirected graph may not have an odd number of vertices with an odd number of neighbors, which would be the case in our model, if 3 people at a party had 3 friends.

This is a basic example of euler's characteristic for connected planar graphs (V-E+F = 2) ? Correct?

The number of vertices of odd degree is always even in an undirected graph. There are three nodes with odd degrees (Three people with three friends). Therefore, no such graph exists.

Seen as their are only 5 people, and 3 of these have 3 friends. So that leaves us with only 2 people left, and they can't be friends with any of the other 3 seen as they only have 3 friends.

In every graph, the sum of degrees of each vertex must be double of the total number of edges. Since (3 3)+(2 2) = 9+4 = 13, and 13 is odd, we can conclude that such a graph is not possible.

Handshaking Lemma