Question 1099798
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(1) Choose the 3 pairs of parents out of the total of 5 pairs to be the ones that will be seated next to their spouses: C(5,3) = 10 {"5 choose 3").
(2)Treat each of the three pairs as units; then the number of ways of arranging those 3 units and the other 4 single parents is 7! = 5040.
(3) In each of the 3 pairs, the two spouses can be arranged in either of 2 ways: 2*2*2 = 8.<br>
All together, the total number of ways of arranging the 10 people is {{{10*5040*8 = 403200}}}.<br>
IMPORTANT NOTE:<br>
I used your given answer to get that result.<br>
As the problem is written, I think the answer should be different -- because in many of the 7! arrangements of the 3 pairs of parents and the other 4 parents (step (2) above), other pairs of spouses will be seated next to each other.  But the problem IMPLIES that EXACTLY 3 pairs of parents are supposed to be seated next to each other.<br>
Counting the number of ways in that case would be far more complicated....