Thirty six identical chairs must be arranged in rows with the same ntimber of chairs in each row. Each row must contain at least three chairs and there must beat least three rows. A row 1S parallel to the front of the room. How many different angementsare possible?
Solution
Case 1: When there is 3 chairs in each row, then total rows = $$\frac{36}{3} = 12$$ (Valid)
Case 2: When there is 4 chairs in each row, then total rows = $$\frac{36}{4} = 9$$ (Valid)
Case 3: When there is 6 chairs in each row, then total rows = $$\frac{36}{6} = 6$$ (Valid)
Case 4: When there is 9 chairs in each row, then total rows = $$\frac{36}{9} = 4$$ (Valid)
Case 5: When there is 12 chairs in each row, then total rows = $$\frac{36}{12} = 3$$ (Valid)
Hence, total possible arrangements = 5
