possible chairs = 5!/(2!*3!)=10 this because chair1, chair2 and chair2, chair1 is the same choice.
total combinations = combination of chairs * combination of tables
150 = 10*combination of tables
combination of tables=15
so x!/(2!*(x-2)!)=15, this simplifies to (x*(x-1))/2= 15 and to x^2-x-30=0
x=6 or x=-5, since you can't have a negative number of tables 6 is the answer (A)
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