Question

In how many ways is it possible to choose a white square and a black square on a chess board so that the squares must not lie in the same row or column?
A.56
B.896
C.60
D.768

Answer 1

Hint: First check number of possibilities for selecting one black. Now remove the whites from the row and column which has the selected black. Now find the number of possibilities for selecting a white. Apply product rules to get the total ways. Total ways is the required result in this question.

Complete step-by-step answer:

Chess Board: A board with $8 \times 8$ dimensions which is formed by alternately placing black and white squares of $\left| x \right|$ dimensions. So, in total it has an area of $64\,c{m^2}$.
This area is formed by $\left| x \right|\,\,c{m^2}$squares. So, total number of squares$= \dfrac{{64}}{1}$.
As they are placed alternately, numbers of black, white are equal. And we know their sum is total. So, $Black + White = 64$
By substituting them as equal, we get their values as:
$2\,Blacks = 64$
By dividing with 2 on both sides, we get it as:
$Black = 32$
There are 8 rows, 8 columns in the board. By simplifying.
By dividing with 8 on both sides of equation, we get:
$\dfrac{{Black}}{8} = \dfrac{{32}}{8}$
By simplifying, we get value of Black in row or column as:
$\dfrac{{Black}}{8} = 4$
From above calculations, we can say all the following statements:
The total number of black squares on the board is 32.
The total number of white squares on the board is 32.
The number of blacks in each row on the board is 4.
The number of whites in each row on the board is 4.
The number of blacks in each row on the board is 4.
The number of whites in each row on the board is 4.

Combinations: It is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter in combination you can select items in any order. Formula is given by ${}^n{C_r} = \dfrac{{n!}}{{\left( {n + 1} \right)!r!}}$
By using the definition of combination we can say that the possibilities of selecting black (be assumed as k) is given by:
$k = {}^{32}{C_1}$
By expanding we can write the value of k in form of:
$k = \dfrac{{32!}}{{30!}}$
By simplifying the above value, we can say value of k to be
$k = 32$.
Now we have to remove whites in the row/column selected black.
$Total - \left( {whites\,\,in\,\,1\,\,row} \right) - \left( {whites\,\,in\,\,1\,\,column} \right) = 32 - 4 - 4$
By simplifying we have more 24 whites to select from.
The possibilities of white selection (be assumed as j) is given by:
$k = {}^{24}{C_1} = \dfrac{{24!}}{{23!}} = 24$
Rule of product: In combinations, the rule of product or, multiplication principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B ways of doing Q work. Given P,Q works can be done at a time. Total number of ways to do both P,Q works are given by $\left( {A.B} \right)$ ways.
By product rule total ways $= j \times k = 24 \times 32$
Total ways are 768.
Therefore option (d) is the correct answer.

Note: Be careful in selecting which of them to be removed. So, calculating whites, blacks in each row carefully. Instead you can select white first and remove blacks from that row, column and then select black. Order doesn’t matter. Anyways you will get the same results.