Question

There are $n$ distinct white balls and $n$ distinct black balls. The number of ways of arranging them in a row so that the neighbouring balls are of different colours is:
A) ${}^{(n + 1)}{C_2}$
B) $(2){(n!)^2}$
C) $\dfrac{{n!}}{2}$
D) None of these

Answer 1

We will take into consideration two cases for the arrangement of balls. We will then find the number of ways of arranging the balls in each case using the given information. Finally, we will add the total number of ways of arranging the balls in the two cases to get the required answer.

Complete step by step solution:
We are given that there are $n$ distinct white balls and $n$ distinct black balls.
We will take two cases of two arrangements by using the given information.

Case 1:
Let the row start with a white ball. The number of ways of selecting the white ball is $n$.
The next ball must be black. Now, there are $n$ ways to select a black ball to fill this position.
The third ball has to be white. The number of ways of selecting this ball is $n - 1$, since one white ball has also been chosen.
Similarly, the fourth ball which is black can be selected in $n - 1$ ways. Hence,
Number of ways of arranging the white balls $= n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 = n!$
Number of ways of arranging the black balls $= n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 = n!$
Total number of ways of arranging $n$ white balls and $n$ black balls in this case $= n! \times n! = {(n!)^2}$

Case 2:
Let the row start with a black ball.
The number of ways of selecting the black ball is $n$.
The next ball must be white. Now, there are $n$ ways to select a white ball to fill this position.
The third ball has to be black. The number of ways of selecting this ball is $n - 1$, since one black ball has also been chosen.
Similarly, the fourth ball which is white can be selected in $n - 1$ ways. Hence,
Number of ways of arranging the black balls $= n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 = n!$
Number of ways of arranging the white balls $= n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 = n!$
Total number of ways of arranging $n$ white balls and $n$ black balls in this case $= n! \times n! = {(n!)^2}$
Therefore, from the above cases, the total number of ways of arranging $n$ distinct white balls and $n$ distinct black balls is ${(n!)^2} + {(n!)^2} = (2){(n!)^2}$.

Thus, the correct answer is option B.

Note:
We have considered two cases above because we do not know which ball is selected for beginning the arrangement. If the colour of the ball which begins the arrangement is mentioned, for example, if the arrangement has to begin with a white ball, then the total number of ways of arranging the balls will be just ${(n!)^2}$. Since this condition is not imposed, we are considering all possible cases.