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
‌8C3.
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
‌7C3. This line of 7 squares, parallel to the central diagonal exists on both sides of it. and hence, the number of ways becomes
‌7C3×2.
Similarly, for the diagonal line having 6 squares, we get
‌6C3×2.
For the diagonal line having 5 squares, we get
‌5C3×2.
For the diagonal line having 4 squares, we get
‌4C3×2.
For the diagonal line having 3 squares, we get
‌3C3×2.
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
(‌8C3+(‌7C3+‌6C3+‌5C3+‌4C3+‌3C3)×2)×2 ⇒(‌+(‌+‌+‌+‌+‌)×2)×2 ⇒(‌+(‌+‌+‌+‌+‌)×2)×2 ⇒(56+(35+20+10+4+1)×2)×2 ⇒(56+140)×2 ⇒392