Question

If four squares are chosen at random on a chess board, The probability that they lie on a diagonal line is $\frac{91k}{{}^{64}{C}_4}.$ The value of $k$ is

Answer 1

Now, for favourable number of cases there are $15$ diagonals in each direction, but only $9$ of them are long enough to contain $4$ squares. specifically, in each direction, there are two diagonals of length $4$, two of length $5$, two of length $6$, two of length $7$ and one of length $8$. So number of cases
$=2{[}^4{C}_4{+}^5{C}_4{+}^6{C}_4{+}^7{C}_4]{+}^8{C}_4$
$=2(1+5+15+35)+70=182$
This is along one diagonal same this for another diagonal.
So, total is $182+182=364$
$\therefore$ Required probability $=\frac{364}{{}^{64}C4}$
Thus, $k=4$.