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?
$\left( {\text{A}} \right)$ $56$
$\left( {\text{B}} \right)$ $896$
$\left( {\text{C}} \right)$ $60$
$\left( {\text{D}} \right)$ $768$

Answer 1

Here we have to choose a white square and a black square as per the given condition and we use a simple combination formula here to solve the problem. Then we multiply the acquired answers to apply both the conditions.
Finally we get the required answer.

Formula used: ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step-by-step solution:
A chess board consists of $64$ squares out of which $32$ are white squares and $32$ are black squares.
First of all, we have to choose $1$ white square out of $32$ white squares in the chess board.
Here we have to use the formula for ${}^n{C_r}$ is equal to $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
For, ${}^{32}{C_1} = \dfrac{{32!}}{{1!\left( {32 - 1} \right)!}}$
On subtracting the denominator term and we get,
$\Rightarrow \dfrac{{32!}}{{1!\left( {31} \right)!}}$
On splitting the factorial term and we get
$\Rightarrow \dfrac{{32 \times 31!}}{{1 \times 31!}}$
Cancel the same term,
$\Rightarrow 32$
Also, we have to know each white square consists of $8$ black squares lying in the same column or row.
So we can write it as, $32 - 8 = 24$ black squares.
Next, we have to choose $1$ black square out of the remaining $24$ black squares in the chess board.
Here we have to find out it as by using the combination formula ${}^{24}{C_1}$
$\Rightarrow \dfrac{{24!}}{{1!\left( {24 - 1} \right)!}}$
Let us subtract the numerator term and we get
$\Rightarrow \dfrac{{24!}}{{1!\left( {23} \right)!}}$
On splitting the factorial we get,
$\Rightarrow \dfrac{{24 \times 23!}}{{1 \times 23!}}$
Cancel the same term and we get,
$\Rightarrow 24$
Now, we have to found that both the white square and black square.
If we choose the White Square and black square, then we have to multiply both the obtained values.
So, number of ways $= {}^{32}{C_1} \times {}^{24}{C_1}$
Putting the finding values and we get,
$\Rightarrow 32 \times 24$
Let us multiply the terms and we get
$\Rightarrow 768$ Ways.
Therefore, it is 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 in $768$ ways.

Hence, the correct answer is option D.

Note: If the number of combination of $n$ different things and we take $r$ at a time allowing repetitions, we can write it as ${}^{n + r - 1}{C_r}$
The number, if we select of $r$ objects out of $n$ identical object is equal to $1$