Holmes wrote:

*Lateral Thinking VII*

You and your spouse invite four other couples to a party. During the course of the conversation, it is discovered that, prior to the party, each person except you was acquainted with a different number of the people present. Assuming the acquaintance relationship is symmetric (i.e., if you are acquainted with someone, that person is also acquainted with you), then how many people did your spouse know prior to the party? How many people did you know?

Good Luck! However, I'm sure you can get it.

Let me do VII first. I found two approaches.

[spoiler]

*The situation*: The guests and your spouse make up nine people who are acquainted with a unique number of people, ranging from 1 to 9. I have included partners as acquaintances, although the arguments below still hold if I preclude them, in which case we would be talking about the range 0 to 8 for the number of acquaintances. "You", the host, have an equal number of acquaintances as one of the guests. What is that number?

*The table*: The following table labels everyone present with Letters A to J and shows all combination of acquaintances. "a" denotes acquaintance, "p" partner (who is also an acquaintance), "h" hosts, and the number on the right diagonal is the number of acquaintances the person in that row/column has. It is the only distribution of acquaintances possible, because the labels are interchangeable and we are initially not assuming anything about who is who's partner.

*Determining "you"*: Filling in the table with "a"s from the top with person A having 9 acquaintances, person B having 8, an so on, we arrive at persons E and F having five acquaintances each. This is unavoidable, hence you, the host, must be either E or F, and be acquainted with exactly 5 people, including your spouse.

*Determining your spouse*: J is acquainted with only one person, A. Therefore A is J's spouse. I is acquainted with two people, A and B, but since A is already taken, B must be I's spouse. Similarly, we can determine the other pairings, H-C, G-D, and find that the host pair is E-F, and both are acquainted with exactly 5 people including themselves. Changing the respective "a"s to "p"s and "h"s results in the table below.

[tt] | J | I | H | G | F | E | D | C | B | A |

----+----+----+----+----+----+----+----+----+----+----+

A | p | a | a | a | a | a | a | a | a | 9

----+----+----+----+----+----+----+----+----+----+

B | | p | a | a | a | a | a | a | 8

----+----+----+----+----+----+----+----+----+

C | | | p | a | a | a | a | 7

----+----+----+----+----+----+----+----+

D | | | | p | a | a | 6

----+----+----+----+----+----+----+

E | | | | | h | 5

----+----+----+----+----+----+

F | | | | | 5

----+----+----+----+----+

G | | | | 4

----+----+----+----+

H | | | 3

----+----+----+

I | | 2

----+----+

J | 1

----+[/tt]

[/spoiler]

Another way of arriving at how many acquaintances "you" as the host have is....

[spoiler]

...the following iterative sequence of arguments, where N=10 and n=1,2,3,...,9:

*(This sentence will seem a bit obscure...)*If there were two people having N-n acquaintances, there would be exactly N-n+1 people having at least n+1 acquaintances. That is not possible, because exactly n people have at most n acquaintances.

*(But the following sentences should make more sense...)*n=1: If there were two people having nine acquaintances, there would be exactly ten people having at least two acquaintances. That is not possible, because exactly one person has at most one acquaintance.

n=2: If there were two people having eight acquaintances, there would be exactly nine people having at least three acquaintances. That is not possible, because exactly two people have at most two acquaintances.

[tt] .

.

.[/tt]

Only for

n=5 the logic does not apply:

If there were two people having five acquaintances, there would be exactly six people having at least five acquaintances. That

*is* possible because two of the six people that have at most five acquaintances, have exactly 5 acquaintances (the supposition in previous sentence), leaving four people with at most four acquaintances (which is ok - cf. consequence in previous sentence).

For

n<5 the sequence continues as before.

Therefore, the only number of acquaintances that can be doubly filled is 5, hence "you", the host, have 5 acquaintances. To find how many acquaintances your spouse has, proceed as in the first solution.

[/spoiler]