Question

The 9 horizontal and 9 vertical lines on an 8 \[ \times \] 8 chess-board form r rectangles and s squares. Then, the ratio s : r in its lowest terms is
A. \[\dfrac{1}{6}\]
B. \[\dfrac{{17}}{{108}}\]
C. \[\dfrac{4}{{27}}\]
D.None of the above

Answer 1

Hint: Choose rectangles using combination and find the number of rectangles. Find the number of squares using the formula of sum of squares of number and then finally find the ratio s : r.

Complete step-by-step answer:
Firstly, we have to select any two horizontal and vertical lines out of 9 horizontal lines and vertical lines respectively for making rectangle or square. For choosing any two lines we have to use a combination.
\[\therefore \]Number of rectangles (r) formed are \[{}^9{C_2} \times {}^9{C_2}\]
 \[ = \dfrac{{9!}}{{2! \times 7!}} \times \dfrac{{9!}}{{2! \times 7!}}\]
 \[ = \dfrac{{9 \times 8}}{2} \times \dfrac{{9 \times 8}}{2}\]
 \[ = 36 \times 36\]
\[\therefore \] r \[ = 36 \times 36\]
Now, for squares, there will be 8 \[ \times \] 8 = 64 squares of size 1 \[ \times \] 1, 7 \[ \times \] 7 = 49 squares of size 2 \[ \times \] 2 and so on.
Thus, number of squares will be \[{8^2} + {7^2} + {6^2} + {5^2} + {4^2} + {3^2} + {2^2} + {1^2}\] .
Sum of squares of numbers is \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] i.e. \[\dfrac{{8\left( {8 + 1} \right)\left( {16 + 1} \right)}}{6} = 12 \times 17\]
\[\therefore \]s \[ = 12 \times 17\]
Now, taking the ratio of s : r
 \[ = \dfrac{{12 \times 17}}{{36 \times 36}}\]
 \[ = \dfrac{{17}}{{108}}\]
Thus, the ratio s : r is equal to \[17:108\] .
Hence correct answer is B.

Note:

Size of square	Numbers of squares
1 \[ \times \] 1	8 \[ \times \] 8 = 64
2 \[ \times \] 2	7 \[ \times \] 7 = 49
3 \[ \times \] 3	6 \[ \times \] 6 = 36
4 \[ \times \] 4	5 \[ \times \] 5 = 25
5 \[ \times \] 5	4 \[ \times \] 4 = 16
6 \[ \times \] 6	3 \[ \times \] 3 =9
7 \[ \times \] 7	2 \[ \times \] 2 = 4
8 \[ \times \] 8	1 \[ \times \] 1 = 1