Question

There are $10$ red balls of different shades and $9$ green balls of identical shades. Then, the number of ways to arrange them in row so that two green balls are together, is

• A. $(10!{)}^{11}{P}_9$
• B. $(10!{)}^{11}{C}_9$
• C. $10!$
• D. $10!9!$

Answer 1

The correct option is B $(10!{)}^{11}{C}_9$

red balls can be arranged in any order, as they are of different shades.

$\Rightarrow 10!$

there are 11 blanks and 9 green balls to fill them.

${\Rightarrow }^{11}{C}_9$

$\therefore (10!{)}^{11}{C}_9$