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(8!)${}^2$
• C. 4!
• D. (4!)${}^3$

Answer 1

Answer: (B)

2(8!)${}^2$

The four different white balls can be arranged in$4!$ ways.

Now we need to arrange black balls

If $1,2,3,4$ are white balls we have arranged black balls in $A,B,C,D$

No. of ways $8!$

However, it can be done as B C D E also

So, total ways $=2\times 8!$

Total ways : $2\times (8!{)}^2.$

Hence, the answer is $2\times (8!{)}^2.$