Question

The number of ways of selecting three squares on a chessboard so that all three be on a diagonal line of the board or parallel to it is

A) 196

B) 126

C) 252

D) 392

Answer 1

Hint: Use the fact that the diagonal line of the board has 8 squares and the number of ways of selecting 3 squares out of it is given by ${}^{8}{{C}_{3}}$. The line next to the diagonal on either side of the chess board has 7 squares and the number of ways of selecting 3 squares out of it is given by ${}^{7}{{C}_{3}}$. Now, this needs to be counted twice as there is one diagonal line of 7 squares on each side of the 8-square diagonal line. Similarly, keep counting the number of squares on each side of the diagonal moving on to 6, 5, 4 and 3 squared diagonals. Take this number and double it as there are two central diagonals of 8 squares each in the middle of the chess board. The number so obtained will be the required number of ways.

Complete step-by-step answer:

First let's 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 ${}^{8}{{C}_{3}}$.

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 ${}^{7}{{C}_{3}}$. This line of 7 squares, parallel to the central diagonal exists on both sides of it and hence, the number of ways becomes ${}^{7}{{C}_{3}}\times 2$.

Similarly, for the diagonal line having 6 squares, we get ${}^{6}{{C}_{3}}\times 2$.

For the diagonal line having 5 squares, we get ${}^{5}{{C}_{3}}\times 2$.

For the diagonal line having 4 squares, we get ${}^{4}{{C}_{3}}\times 2$.

For the diagonal line having 3 squares, we get ${}^{3}{{C}_{3}}\times 2$.

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

\[ \left( {}^{8}{{C}_{3}}+\left( {}^{7}{{C}_{3}}+{}^{6}{{C}_{3}}+{}^{5}{{C}_{3}}+{}^{4}{{C}_{3}}+{}^{3}{{C}_{3}} \right)\times 2 \right)\times 2\]

\[\Rightarrow \left( \dfrac{8!}{3!\left( 8-3 \right)!}+\left( \dfrac{7!}{3!\left( 7-3 \right)!}+\dfrac{6!}{3!\left( 6-3 \right)!}+\dfrac{5!}{3!\left( 5-3 \right)!}+\dfrac{4!}{3!\left( 4-3 \right)!}+\dfrac{3!}{3!\left( 3-3 \right)!} \right)\times 2 \right)\times 2\]

\[\Rightarrow \left( \dfrac{8!}{3!5!}+\left( \dfrac{7!}{3!4!}+\dfrac{6!}{3!3!}+\dfrac{5!}{3!2!}+\dfrac{4!}{3!1!}+\dfrac{3!}{3!0!} \right)\times 2 \right)\times 2 \]

\[\Rightarrow \left( 56+\left( 35+20+10+4+1 \right)\times 2 \right)\times 2\]

\[\Rightarrow \left( 56+140 \right)\times 2\]

\[ \Rightarrow 392 \]

Thus the total number of ways comes out to be 392, and the correct answer is an option (D).

Note: It is important to keep in mind that the total number has to be multiplied to 2 at appropriate places to account for the two identical diagonals on each side of the central diagonal and the two central diagonals present in the chess board.