Md.Nazrul Islam wrote:one single person and two couple are to be seated at random in a of raw five chairs ,what is the probability that neither of the couples sits together in adjacent chairs .
Let's say that we have 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 is to SUBTRACT THE OVERLAP.
When we count the arrangements in which AB sit together and those in which CD sit together, the arrangements in which
both AB and CD sit together -- the OVERLAP -- gets counted twice.
Thus, we need to SUBTRACT THE OVERLAP -- the arrangements in which both AB and CD sit together -- so that these arrangements are not double-counted.
Total arrangements = 5! = 120
AB together:
Number of ways to arrange the 4 elements AB, C, D, and E = 4! = 24.
Since AB can be reversed to BA, we multiply by 2:
2*24 = 48.
CD together:
Number of ways to arrange the 4 elements CD, A, B and E = 4! = 24.
Since CD can be reversed to DC, we multiply by 2:
2*24 = 48.
Both AB and CD together:
Number of ways to arrange the 3 elements AB, CD, and E = 3! = 6.
Since AB can be reversed, CD can be reversed, and both AB and CD can be reversed, 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, P(neither couple sit together) = 48/120 = 2/5.
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