There are $(n + 1)$ white and $(n + 1)$ black balls,with each set numbered $1$ to $n + 1$. The number of ways in which the balls can be arranged in a row so that no two adjacent balls are of the same colour is

  • A
    $(2n + 2)!$
  • B
    $(2n + 2)! \times 2$
  • C
    $(n + 1)! \times 2$
  • D
    $2 \{(n + 1)!\}^2$

Explore More

Similar Questions

If a number is chosen at random from the four-digit numbers formed by using the digits $0, 1, 2, 3, 4, 6$ without repetition,then the probability that it is divisible by $4$ is

$A$ test containing $3$ objective-type questions is conducted in a class. Each question has $4$ options and only one option is the correct answer. No two students of the class have answered identically,and no student has written all correct answers. If every student has attempted all the questions,then the maximum possible number of students who have written the test is:

$A$ question paper contains $4$ questions,each having $4$ alternative answers. The number of ways that a candidate can answer one or more questions is

There are $n$ white and $n$ black balls marked $1, 2, 3, \ldots, n$. The number of ways in which we can arrange these balls in a row so that neighbouring balls are of different colours is

The number of three-digit numbers $\overline{abc}$ such that the arithmetic mean of $b$ and $c$ is equal to the square of their geometric mean is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo