Mission2012 wrote:Two couples and one single person (5th wheel) go to see a movie and find a line of 5 available seats. They randomly fill in those 5 seats. What is the probability that no couples sit next to each other?

(A)
1/4

(B)
1/2

(C)
1/5

(D)
2/5

(E)
3/4

Let the 5 people be couple AB, couple CD and lonely boy E.

Let:
T = total possible arrangements
AB = arrangements in which AB sit in adjacent seats
CD = arrangements in which CD sit in adjacent seats
AB+CD = arrangements in which BOTH COUPLES sit in adjacent seats
N = arrangements in which NEITHER COUPLE sits in adjacent seats.

T = AB + CD - (AB+CD) + N

When we count AB and then CD, the OVERLAP -- the arrangements in which BOTH COUPLES sit in adjacent seats (AB+CD) -- is counted TWICE.
So that we don't double-count the overlap, AB+CD must be subtracted from the total.

T = number of ways to arrange 5 elements = 5! = 120

AB:
Here, 4 elements are to be arranged: C, D, E, and couple AB.
Number of ways to arrange 4 elements = 4! = 24.
Since AB can be reversed to BA, we multiply by 2:
2*24 = 48.

CD:
Applying the same reasoning used for AB, we get:
2*24 = 48.

AB+CD:
Here, 3 elements are to be arranged: E, couple AB, and couple CD.
Number of ways to arrange 3 elements = 3! = 6.
Within the two couples, there are 4 ways to arrange the spouses themselves:
AB - CD
BA - CD
AB - DC
BA - DC.
Thus, we multiply by 4:
4*6 = 24.

Plugging these values into the equation above, we get:

120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48

Thus:
N/T = 48/120 = 2/5.

The correct answer is D.

Mitch ,I for some reason got the following :

Possible combinations couples together: 3!*2*2 ...considering that a couple is one person and there are 3 seats. (120-48)/120 = 72/120

Not sure which combinations am I missing