Chess and probability

I have a nice sol. - In a group of 4 squares of a chessboard , one can form 4 L. Take 1 square box as $1\times 1$ dimension So a group of 4 boxes will make $2\times 2$ dimension . In a Chessboard there are 49 squares = (8-1)(8-1) So total L's are 49*4 = $\boxed{196}$

Take bottom left corner of the chessboard as origin, bottom edge as positive x-axis and left edge as positive y-axis. Observe that if we take two diagonally opposite squares, they would specify a $2 \times 2$ square. In this square, four $L$ shapes can be realized. If we count all such $2 \times 2$ squares, then four times that should give us the value of $A$ . However, such $2 \times 2$ square can be specified in two different ways since it has two diagonals. Let us do a trick to avoid repeating the counting of $L$ shapes. In each such square, draw an arrow along the main diagonal which is parallel to line $x=y$ . We will count squares using this arrow only. Along any diagonal of length $n$ of the chessboard with $n>1$ , we can draw $n-1$ arrows. This will mean that along this diagonal, we can specify $4(n-1)$ squares that form $L$ shape. Along the main diagonal there are $7$ such arrows while along the other diagonals, there are $42$ arrows. Thus there are $42+7=49$ arrows and hence $A = 4 \times 49 = \boxed{196}$ .