Question

There are 10 white and 10 black balls marked 1, 2, 3,…….10. The number of ways in which we can arrange these balls in a row, in such a way that, neighboring balls are of different colors is
A) \[10!9!\]
B) \[20!\]
C) \[{\left( {10!} \right)^2}\]
D) \[2{\left( {10!} \right)^2}\]

Answer 1

Hint:
Here, we have to find the number of ways that the balls are arranged alternately in a row. This can be found out by using the concept of factorial. First, we will find the number of ways to select the first ball as black. Then we will find the number of ways to select the first ball ss white. We will add these to find the total number of ways. Factorial is defined as the number of ways of arrangement of the objects in an order.

Complete step by step solution:
We are given 10 white balls and 10 black balls marked 1, 2, 3,…………., 10.
First, we will find the number of ways in which the first ball is black.
Black balls are arranged in 10 different ways and white balls are arranged in 10 different ways.
Number of ways in which the first ball is black\[ = 10! \times 10!\]
Now, we have to find the number of ways in which the first ball is white.
White balls are arranged in 10 different ways and black balls are arranged in 10 different ways.
Number of ways in which the first ball is white\[ = 10! \times 10!\]
Now, we can find the total number of ways by adding both the cases. Therefore, we get
 Total Number of ways\[ = 10! \times 10! + 10! \times 10!\]
Adding the terms, we get
\[ \Rightarrow \] Total Number of ways\[ = 2\left( {10! \times 10!} \right)\]
\[ \Rightarrow \] Total Number of ways\[ = 2{\left( {10!} \right)^2}\]
Therefore, the number of ways of arranging the balls alternately is \[2{\left( {10!} \right)^2}\] and thus

Option (D) is correct.

Note:
We know that the number of ways of arrangements of letters, numbers, etc in a finite order is known as permutations. The number of ways of selecting items where the order is not important is known as combinations. The concept of factorial is used in both the permutations and combinations. We should also remember that the first ball may be a blackball or a white ball.