15. Michela has drawn a rectangular table of dimensions $2 \times 100$. She wants to place 99 coins, each in one cell of the table, in such a way that there are no pairs of cells with a common side both occupied by a coin. In how many different ways can Michela place the 99 coins?
(A) 200
(B) 396
(C) 402
(D) 400
(E) 202
(15) The correct answer is $(B)$.

For every possible arrangement compatible with Michela's requirements, there will be one and only one column left empty. If this column is the first (or the last), there are exactly 2 possible arrangements, determined by which cell in the adjacent column contains a coin. If, instead, the empty column is neither the first nor the last (thus one of the remaining 98), there are 4 possible arrangements, determined by the 4 possible placements of two coins in the cells of the two adjacent columns. In total, the possible arrangements will be $2 \cdot 2 + 4 \cdot 98 = 396$. Problem proposed by Matteo Rossi.