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 colours is

$$A.10!9! \\\ B.20! \\\ C.{\left( {10!} \right)^2} \\\ D.2{\left( {10!} \right)^2} \\\$$

Answer 1

The balls are to be arranged alternately. We must find the number of ways balls of each colour can be arranged in alternate positions and take the product. This can be found using permutations. We must also consider that the 1st ball can be either white or black. This factor also must be multiplied to get the final answer.

Complete step-by-step answer:
Here we are asked to find the number of ways the balls of two different colours can be arranged such that two adjacent balls don’t have the same colour. This also means that the black and white balls must be placed alternately. According to the question there are 10 black balls to be placed alternately with 10 white balls. The number of ways of arranging the 10 black balls is given by 10!. The number of ways of arranging 10 white balls is also given by 10! . But here, the 1st ball can be either white or black. So the total number of ways of arranging 10 black balls and 10 white balls alternatively is given by the product $2 \times 10! \times 10!$.
So the number of ways the balls can be arranged such that neighboring balls have different colour is $2{\left( {10!} \right)^2}$
Hence the correct answer is option D.

Note: Visualizing the problem will help us to understand it better. A common error is that the possibility of selecting the 1st ball is not considered. The concepts of permutations and combinations required in this problem