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