The square shown is divided into five cells. How many paths can be drawn that start at any cell, move only to adjacent cells and visit each of the five cells exactly once?

A scrutiny of the given figure reveals that there are 5 cells A,B,C,D and E of which the first 4 cells are at the 4 corners and the cell E is in the center. We can start with any cell. Let us choose first the corner cells to start determining the paths which, starting from a cell, move to adjacent cells and visit each cell only once.
Let us start with the cell A. This cell has 3 adjoining cells, B,D and E. Let us start from A, followed by B. The requiblack paths are ABCDE, ABCED, ABECD, ABEDC . Thus, there are 4 paths which satisfy the given condition. Similarly, If we start from A, followed by D, there will be 4 paths which satisfy the given condition. Next, if we start from A, followed by E, the paths AEBCD, AEDCB only satisfy the given condition. Further, we cannot have any path starting from A and followed by C as it is not an adjoining cell. Thus, there can be 4+4+2=10 paths starting from A which satisfy the given condition.
Similarly, there will be 10 paths , if we start from any of the cells B,C or D. Thus, there will be 4*10 = 40 paths which satisfy the given condition if we start from A,B,C or D.
For the cell E, there are 4 adjoining cells A,B,C and D. If we start from E followed by A, the paths which satisfy the given condition are EABCD, EADCB. Similarly, if we start from E followed by B,C or D, there will be 2 paths which satisfy the given condition. Thus, there will be 4*2 = 8 paths which satisfy the given condition if we start from the cell E.
On adding up, there are 40 +8 = 48 paths which, starting from a cell, move to adjacent cells and visit each cell only once.