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My approach was
Positions 1 2 3 4 5 6
case 1 J F F F F F => 5 different positions possible
2: J F F F F => 4 possible positions
3: J F F F => 3 positions
4: J F F => 2
5: J F => 1
A total of 5+4+3+2+1 = 15 positions possible for just 2 of them
For each of this the rest of them can be arranged in 4! ways (4 positions 4 people)
Hence 15 * 24 = 360
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