Another Handshake Problem

[Divinity]

Each person shakes the hand of a different number of people, except possibly Mrs. Delphine. The only possible numbers given the parameters of the problem are: 0, 1, 2, 3, 4, 5.

Married Couple #1: A&B
Married Couple #2: C&D
Married Couple #3: E&F
Mrs. Delphine: G

Pair the highest number with the lowest number because in order for A to shake 5 hands A must shake the hands of each person (including Mrs. Delphine), meaning none of the other guests can shake 0 hands because A has already shook their hand. Each couple then is paired off, highest and lowest again because they can't shake hands with each other. If C = 1, then D = 4, because E and F can't shake each other's hands, and cannot reach 4 because E and F are a couple and only have A & D. Adding Mrs. Delphine or not only makes 3.

A=5
B=0
C=1
D=4
E=3
F=2

A shakes the hand of C, D, E, F, and Mrs. Delphine (5)
C shakes the hand of A (1)
D shakes the hand of A, E, F, and Mrs. Delphine (4)
E shakes the hand of A, D, and Mrs. Delphine (3)
F shakes the hand of A and D (2)

Mrs Delphine shakes three hands.
Total number of handshakes is 15

[Whateverist]

I basically worked it out with a table like Clueless Morgan posted above, and got the answer as 3 as well.

Hey Tiberius.  Have you had a look at Rob's abortion of a probability problem, drawing names from hats?  As I got close enough to appreciate what all is required, it strikes me as just a lot of brute counting and calculating with no likely elegance in the solution.  Do you know the problem?  If so can you say whether you think its solution is likely to warrant the effort?