Since all the men like 1 better, 1 gets to choose. (A, 1) is the natural choice.
Does (B, 2) lead to an instability?
B would prefer 1, but 1 already has a better match than B.
2 would prefer A, but again A already has a better match than 2
So this is not an instability
Case 2 from the book
M: {A, B}
W: {1, 2}
A: 1, 2
B: 2, 1
1: B, A
2: A, B
No one choice is clear.
S = { (A,1), (B, 2) } is valid
A prefers 1 and B prefers 2, so they would not switch in any situation.
1 and 2 don't get their choices, but swapping would not make A and B more happy.
So S is stable.
S = { (A, 2), (B, 1)}
This is the opposite of the first set, but the same holds
1 and 2 are very happy and would not switch.
A and B might want to switch but their prospective partners lot would not improve
So this is a stable match as well.
S = { (A, 1) , (A, 2)}
This is not 1:1 or perfect so the match is not stable.
Consider the following
$M=\{A,B,C,D\}$
$W=\{1,2,3,4\}$
A: 1, 2, 3, 4
B: 1, 3, 2, 4
C: 4, 2, 3, 1
D: 4, 3, 2, 1
1: A, B, C, D
2: A, B, C, D
3: A, B, C, D
4: D, C, A, B
Could we do any pairing other than (A,1) and have a stable match
Assume (A,n), n≠ 1 is a match.
Then assuming S is perfect, (m,1), m≠ A is also a match.
But
A prefers 1 to all other n,
and 1 prefers A to all other m
so (A,n) and (m,1) are instabilities.
Therefore (A,1) ∈ S
You could also argue that (D,4) ∈ S
How about B: 3, 2 and C: 2, 3 and 2: B, C and 3: B, C
Since B prefers 3, and Both 2 and 3 prefer B, the match (B, 3) must be made
Otherwise in (B, 2) and (C, 3) the match (B, 3) represents an instability.
I think this is the missing case from all of the two person scenarios.
Is the following stable?
M: {A, B, C}
W: {1, 2, 3}
A: 1, 2, 3
B: 2, 1, 3
C 1, 2, 3
1: A, B, C
2: A, C, B
3: A, C, B
S = { (A, 1), (B, 2), (C, 3)}
(C, 2) would make both C (who prefers 2 to 3) and 2 (who prefers C to B) happier.
Therefore the match S is not stable.
So S = {(A,1), (B,3), (C,2)} is stable.