Consider a  $6\times 6$  chessboard. Then match the following lists:

	List -I		List -II
A)	Number of rectangles	p)	${}^{10}C_5$
B)	Number of squares	q)	441
C)	Number of ways three squares can be selected if they are not in same row or column.	r)	91
D)	In how many ways eleven ‘‘+’’ sign  Can be arranged in the squares if no  Row remains empty.	s)	2400

a) Number of rectangles is equal to number of ways we can select two vertical lines and two horizontal lines. Total number of ways is  ${}^{7}C_2\times {}^{7}C_2=441$ .
 b) If the square is of 1 sq. units such as a6, then we have such $6\times 6=36$   squares.
If the square is of 4 sq. units like the shaded region of the squares a1,a2,b1,b2, then we have such 5 squares in the belt formed by rows 1 and 2. Similarly, we have 4 more belts 23,34,45, and 56. Hence, there are $5\times 5=25$   such squares.
Similarly, we have  $4\times 4,3\times 3,2\times 2,1\times 1$  squares if increasing sizes. 
Hence, total number of squares is $1+4+9+16+25+36=91$
c) The first square can be selected in 36 ways. If one such square a1 is selected, we are left with 25 squares; second square cannot be selected from row 1 and column a. if second square is c2, we are left with 16 squares, from which third square can be selected, e.g., b4.
Hence, number of ways of selections is  $36\times 25\times 16$ . But in this one-by-one type of selection order of selection is also considered. Hence, actual number of ways is  $(36\times 25\times 16)/3!$= 2400.
d) Given number of ways is equivalent to selecting 11 squares from 36 squares if no row remains empty.
Suppose $x_1,x_2,x_3,x_4,x_5,x_6$   be the number of squares selected from the 1st, 2nd, 3rd, 4th, 5th and 6th row. 
Then we must have  $x_1+x_2+x_3+x_4+x_5+x_6=11$  (where $1\le x_i\le 6$  ).
The number of positive integral solutions of the above equations is  ${}^{11-1}C_{6-1}={}^{10}C_5$ .