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  • How many different methods are there to select two white squares and two black squares on a chessboard so they cannot be placed in the same row or column?  <div class='qna-option

How many different methods are there to select two white squares and two black squares on a chessboard so they cannot be placed in the same row or column?

 

Option: 1

124096


Option: 2

136896 
 


Option: 3

143786

 


Option: 4

152136


Answers (1)

best_answer

Since there are 8 rows and 8 columns on a chessboard, the total number of squares is,

8\times 8=64

From this, the number of white squares is,

\frac{64}{2}=32

Thus, the total number of white squares is 32 and the total number of black squares is 32.

Therefore ways to choose 2 white square out of 32 white squares is given by a combination formula,

\mathrm{\begin{aligned} &{ }^{32} C_2=\frac{32 !}{2 ! 30 !}\\ &{ }^{32} C_2=\frac{32 \times 31}{2}\\ &{ }^{32} C_2=496 \end{aligned}}

Now we have to select the black squares. Given that the black and the white squares cannot be placed in the same row or column.

So, canceling out the black and the white squares from 1 row and 1 column is,

N = 32 - 8 = 24

The number of ways to choose 2 black squares from the remaining squares is given by,

\mathrm{\begin{aligned} { }^{24} C_2 & =\frac{24 !}{2 ! 22 !} \\ { }^{24} C_2 & =\frac{24 \times 23}{2} \\ { }^{24} C_2 & =12 \times 23 \\ { }^{24} C_2 & =276 \end{aligned}}

Thus, the total number of possible selections for a white and a black square that exclude squares from being placed in the same row or column is given by,

\mathrm{\begin{aligned} &{ }^{32} C_2 \times{ }^{24} C_2=496 \times 276\\ &{ }^{32} C_2 \times{ }^{24} C_2=136896 \end{aligned}}

Therefore, the total number of possible ways is 136896 ways.

 

 

Posted by

Nehul

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