Question

There are four different white balls and four different black balls. The number of ways that balls can be arranged in a row so that white and black balls are placed alternately is

• A. $(4!{)}^2$
• B. $2(4!{)}^2$
• C. $4!$
• D. $(4!{)}^3$

Answer 1

Answer: (B)

$2(4!{)}^2$

We have to arrange 4 different white balls and 4 different black balls as shown below

(i) WB WB WB WB

(ii) BW BW BW BW

This can be done in $4!\times 4!+4!\times 4!=2(4!{)}^2$ ways.