The ABO and MN blood groups are given below for four sets of parents (1 to 4) an

The statement:
"Recall that the ABO blood group has three alleles, IA, IB, and i. The MN blood group has two codominant alleles, M and N."
is very helpful.

We should be able to deduce:
In the ABO typing system,
Type A is iAiA or iAi
Type B is iBiB or iBi
Type AB is iAiB only
Type O is ii only

In the MN typing system,
Type M is MM only
Type MN is MN only
Type N is NN only

Let's use this information:
Mother1: ii; MM only
Father1: iBiB or iBi; MM only
Mother 2: iBiB or iBi; NN only
Father 2: iBiB or iBi; NN only
Mother 3: iAiB; MN only
Father 3: iBiB or iBi; MN only
Mother 4: iAiA or iAi; NN only
Father 4: iBiB or iBi; MN only

Remember that each parent donates one allele from each blood typing.

Child a (B; M):
iBiB or iBi; MM only
Parents 1 are included.  Mother 1 donates i; M and Father 1 donates iB; M
Parents 2 are excluded.  While they could have accounted for the ABO blood type, two N parents will have all N children (NN x NN = all NN)
Parents 3 are included.  Mother 3 donates iB; M and Father 3 donates iB; M
Parents 4 are excluded.  While they could have accounted for the ABO blood type, an N parent and an MN parent will have only N and MN children (NN x MN = NN or MN)

Child b (O; M):
ii only; MM only
Parents 1 are included.  Mother 1 donates i; M and Father 1 donates i (remember he can be iBi or iBiB); M
Parents 2 are excluded.  While they could have accounted for the ABO blood type, two N parents will have all N children (NN x NN = all NN)
Parents 3 are excluded.  Mother 3 is iAiB.  She can only produce A, B, or AB offspring.
Parents 4 are excluded.  While they could have accounted for the ABO blood type, an N parent and an MN parent will have only N and MN children (NN x MN = NN or MN).

Child c (AB, MN):
iAiB; MN
Parents 1 are excluded.  Neither has an iA allele.
Parents 2 are excluded.  Neither has an iA allele.
Parents 3 are included.  Mother 3 donates an iA and M or N.  Father 3 donates an iB and the M or N (the opposite of the mother's MN allele).
Parents 4 are included.  Mother 4 donates iA; N.  Father 4 donates iB; M.

Child d (B, N):
iBiB or iBi; NN only
Parents 1 are excluded.  While they could have accounted for the ABO blood type, two M parents will have all M children (MM x MM = all MM)
Parents 2 are included.  Both mother 2 and father 2 could have donated either i or iB (as long as one or both parents donates an iB, as ii would be an O offspring);N
Parents 3 are included.  Mother 3 donates iB; N and Father 3 donates either iB or i;N.
Parents 4 are included.  Mother 4 donates i;N.  Father 4 donates iB;N.

I hope this all makes sense.  I tried to make it as verbose as possible so you could learn from it.

1=b : 2=d : 3=c : 4=a

Couple 1: ii MM x IBi MM
Possible children: IBi MM or ii MM -- B,M or O,M

Couple 2: IBi NN x IBi NN
Possible children: IBIB NN or IBi NN; ii NN -- B,N or O,N

Couple 3: IAIB MN x IBi MN
Possible children: IAIB; IAi; IBi; (with MM or MN or NN) -- AB,M; AB,MN; AB,N; AM; A,MN; A,N; B,M; B,MN; B,N

Couple 4: IAi NN x IBi MN
Possible children: IAIB, IAi, IBi, ii (MN or NN) -- AB,MN; AB,N; A,MN; A,N; B,MN; B,N; O,MN; O,N

Children a-d:
a: B,M - could be from couple 1 or 3
b: O,M - could be from couple 1 only
c: AB,MN - could be from couple 3 or 4
d: B,N - could be from couple 2 or 4

By process of elimination:
Couple 1: b
Couple 2: a
Couple 3: c
Couple 4: d

That answer is incorrect.  Couple 3 could also be the parents of child d. (AB;MN x B;MN = BN from iB;N maternal and iB;N paternal = iBiB;NN (child d)).

Also, from the way the question is worded, the purpose is not to pair the parents to children but to state all possibilities and exclusions.  Given this new piece of information, however, assuming that the question wants an actual pairing, the correct pairing would be:

Children a-d:
a: B,M - could be from couple 1 or 3
b: O,M - could be from couple 1 only
c: AB,MN - could be from couple 3 or 4
d: B,N - could be from couple 2, 3, or 4

By process of elimination:
Couple 1: b
Couple 2: d
Couple 3: a
Couple 4: c

Steps to this conclusion:
b can be 1 only, therefore we can eliminate 1 from a and deduce that a is 3.  We can then eliminate 3 from c and d, and therefore, only 4 can be c and this leaves 2 for d.

Still think that's more logic than genetics, though.  This assumes that these are the only possible parents and children (not explicitly stated).  I think the inclusion/exclusion statements are the best answer.

thanks

Thx