VITEEE 2016 Paper

First let the count the number of ways of selecting 3 squares on one of the central diagonal of the chess board having 8 squares. This is given by ${}^8C_3$. Next we obtain the number of ways of selecting 3 squares on the next line parallel to this central diagonal, which is the line having 7 squares. This is given by ${}^7C_3$. This line of 7 squares, parallel to the central diagonal exists on both sides of it. and hence, the number of ways becomes ${}^7C_3 \times 2$. Similarly, for the diagonal line having 6 squares, we get ${}^6C_3 \times 2$. For the diagonal line having 5 squares, we get ${}^5C_3 \times 2$. For the diagonal line having 4 squares, we get ${}^4C_3 \times 2$. For the diagonal line having 3 squares, we get ${}^3C_3 \times 2$. Now, all of this needs to be considered twice as we can have a central diagonal from both sides. Thus the total number of ways will come out to be this sum multiplied by 2 . Thus, the total number of ways is given as $({}^8C_3 + ({}^7C_3 + {}^6C_3 + {}^5C_3 + {}^4C_3 + {}^3C_3) \times 2) \times 2$ $\Rightarrow \left( \; \frac{8!}{3!(8-3)!} + \left( \; \frac{7!}{3!(7-3)!} + \; \frac{6!}{3!(6-3)!} + \; \frac{5!}{3!(5-3)!} + \; \frac{4!}{3!(4-3)!} + \; \frac{3!}{3!(3-3)!} \right) \times 2 \right) \times 2$ $\Rightarrow \left( \; \frac{8!}{3!5!} + \left( \; \frac{7!}{3!4!} + \; \frac{6!}{3!3!} + \; \frac{5!}{3!2!} + \; \frac{4!}{3!1!} + \; \frac{3!}{3!0!} \right) \times 2 \right) \times 2$ $\Rightarrow (56 + (35 + 20 + 10 + 4 + 1) \times 2) \times 2$ $\Rightarrow (56 + 140) \times 2$ $\Rightarrow 392$