A park contains 4 loving couples. 3 people are randomly selected. The chance of picking a couple amongst the 3 is... 
A park contains 4 loving couples.  3 people are randomly selected.  The chance of picking a couple amongst the 3 is...

My solution:


*

*1st Move: Select any person 8 ways.  

*2nd Move: Select the person’s mate 1 way  

*3rd Move: Select any person 6 ways.  
Total ways: $8 \cdot 6 \cdot (3!/2!)$ -- where (3!/2!) is the permutation factor.
Sample Space $8P3$ 
Therefore, Probability is:
$$\frac{8 \cdot 6 \cdot (3!/2!)}{8P3}= 0.429$$
However, it seems obvious that my probability cannot be this high. Did I make a mistake somewhere? Thanks stack!

Here is my solution. Consider the number of ways you can select 3 people out of a set of 8.
The total number (in your terms, I believe, is unrestricted) of combinations is:
$$
 ^8C_3
$$
Consider the number of ways to select a set of 3 people with 1 couple.
Every couple selected can have 1 out of the remaining 6 people. There are 4 couples.
Hence, the number of combinations is
$$4 \times 6$$
Hence the resultant probability is
$$
\frac{4\times6}{{8\choose3}} = 0.429
$$
I think your answer is correct.
Being Singaporean, I believe you still have doubts, based on my interaction with Singaporean students. Let me double-confirm (if it's the right way of saying) your answer.
Let's now consider that of the 4 couples, I select any 3 couples. The number of ways is $$^4C_3 = 4$$
Of a set of 3 couples, you can select 1 man or 1 woman each to form a set of 3 people who are not couples. The total number of ways is
$$2^3 = 8$$
Thus, the number of ways to select 3 people who are not couples is:
$$4 \times 8 = 32$$
If you add $32$ and $24$, you get $^8C_3 = 56$, which is the total number of combinations. The answer is indeed correct since the set of 3 either contains a couple or does not (they, combined form the universal set for this question).
