Question

Let there are (n + 1) white and (n + 1) black balls. In each set, the balls are numbered from 1 to (n + 1). If these balls are to be arranged in a row so that consecutive balls are of different colours, then the number of these arrangements is

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

Answer 1

The correct option is B $2\times \left(\right(n+1)!{)}^2$

$2\times (n+1)!\times (n+1)!$