10 people want to sit in a row with only 3 seats.
Obviously they can't, so only 3 will be assigned to the 3 seats, and the other 7
will be left standing. There is a particular couple among the 10 people, say,
Alice and Ben. You want to know the probability that we have one of these
seating arrangements
CXY, XCY, XYC, CXC, XYZ
where C represents either Alice or Ben, and where X, Y, Z are neither Alice
nor Ben.
Each of the 1st 3 seating arrangements can be chosen (2)(8)(7)=112 ways,
so they account for (3)(112)=336 ways.
The 4th case can be chosen (2)(8)(1)=16 ways.
The 5th case can be chosen (8)(7)(6)=336 ways.
That's 336+16+336=688 cases
The total ways any 3 of the 10 can be seated. Choose the 3 to be seated in
(10)(9)(8)=720
Desired probability = 688/720 = 43/45
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I just realized it would be easier to go for the complement event.
CCX, XCC.
Choose the CC 2 ways and X 8 ways. That's 16 ways for CCX and 16 ways for XCC.
So there are (2)(16)=32 unsuccessful ways out of (10)(9)(8)=720 ways, leaving
688 successful ways, out of 720.
Answer: 688/720 = 43/45.
Edwin
