J. Weldon Wolfingham, Jr. will marry one of these women: Adrianna, Bianca, and Claire.
Some clues are furnished hereunder as follows:
1. Of Adrianna and Bianca:
• (a) Either they are both have blue eyes, or neither has blue eyes.
• (b) One has red hair and the other does not.
2. Of Adrianna and Claire:
• (a) Either they both have red hair, or neither has red hair.
• (b) One is 5'11" and the other is not.
3. Of Bianca and Claire:
• (a) One has blue eyes and the other does not.
• (b) One is 5'11" and the other is not.
4. Of the three characteristics  blue eyes, red hair and 5'11":
• (a) If any of the three women has exactly two of the three characteristics, Mr. Wolfingham will marry the one with the least number of characteristics.
• (b) If any of the three women has exactly one of the three characteristics, Mr. Wolfingham will marry the one with the greatest number of characteristics.
Who will Mr. Wolfingham marry?
Clair.
Having made a table of women vs characteristics, you can see that:
the "odd man out" for eye color is Clair,
the "odd man out" for hair color is Bianca,
the "odd man out" for height is again Clair.
But we don't know if the "odd man out"
has the characteristic or does
not have it.
There are 8 possible ways the 3 characteristics could be distributed among the "odd men". For each of the 8 ways, I added to the table, counting the number of characteristics of each woman.
A 3 2 2 1 2 1 1 0
B 2 1 2 2 1 0 3 1
C 1 2 0 2 2 3 2 2
Any scenario in which one woman has 2 characteristics and another woman has 1 characteristic leads to a logical inconsistency, due to condition 4. This allows elimination of 6 of the 8 columns:
A 2 1
B 2 0
C 0 3
Either of these leads to the marital choice of Clair.

Posted by Larry
on 20230301 10:12:21 