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The solution goes as follows:
Aisle seat = 1 and number of occupants are 3 i.e., Archie, Jerry, or Moose
Middle seats = 3
Third row = 1
Now, \(3C1\) ways to fill the aisle seat. Remaining 4 people can fill the remaining 4 seats in 4! ways. However, we need to deduct the cases where Betty and Veronica sit next to each other and there would be 8 (2*2*2) ways for Betty and Veronica to sit together.
These two can sit together in middle row of the front seat only. There can be two possible arrangement as follows:
(B,V,_) and (_,B,V)
These two can switch between each other so total ways become 2*2. Now for this arrangement the remaining two seats i.e., one seat of middle row and one seat of third row can be filled in 2! ways. And therefore, total number of ways would be 2*2*2.
Plugging in the values,
\(3C1\)*(4!-2*2*2) = 3*(24-8) = 48
Aisle seat = 1 and number of occupants are 3 i.e., Archie, Jerry, or Moose
Middle seats = 3
Third row = 1
Now, \(3C1\) ways to fill the aisle seat. Remaining 4 people can fill the remaining 4 seats in 4! ways. However, we need to deduct the cases where Betty and Veronica sit next to each other and there would be 8 (2*2*2) ways for Betty and Veronica to sit together.
These two can sit together in middle row of the front seat only. There can be two possible arrangement as follows:
(B,V,_) and (_,B,V)
These two can switch between each other so total ways become 2*2. Now for this arrangement the remaining two seats i.e., one seat of middle row and one seat of third row can be filled in 2! ways. And therefore, total number of ways would be 2*2*2.
Plugging in the values,
\(3C1\)*(4!-2*2*2) = 3*(24-8) = 48
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