hey_thr67 wrote:One single person and two couples are to be seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs.
A)1/5
B)1/4
C)3/8
D)2/5
E)1/2
OA is D
Let the 5 people consist of couple AB, couple CD, and lonely person E.
Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)
The big idea with overlapping groups is to SUBTRACT THE OVERLAP.
When we count the arrangements in which AB sit together, among those arrangements will be some in which CD also sit together.
When we count the arrangements in which CD sit together, among those arrangements will be some in which AB also sit together.
The result is that the OVERLAP -- the arrangements in which
both AB and CD sit together -- will be counted twice.
Thus, we SUBTRACT THE OVERLAP so that it is not double-counted.
Total arrangements = 5! = 120
AB together:
Here, we're arranging the 4 elements AB, C, D, and E.
Number of ways to arrangement 4 elements = 4! = 24.
Since AB can be reversed to BA, we multiply by 2:
2*24 = 48.
CD together:
Using the same reasoning used for AB, the number of arrangements here = 48.
AB and CD together:
Here, we're arranging the 3 elements AB, CD, and E.
Number of ways to arrange 3 elements = 3! = 6.
Since AB can be reversed, CD can be reversed, and both AB and CD can be reversed -- yielding a total of 4 variations -- we multiply by 4:
4*6 = 24.
Plugging these results into the overlapping groups formula, we get:
120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48
So P(neither couple sitting together) = 48/120 = 2/5.
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