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 a row so that green balls are together is

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

Answer 1

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

Consider $R_1,R_2,.....R_{10}$ are $10$ red balls ………

$R_1$-$R_2$-$R_3$-...…..-$R_{10}$

There are $11$ blanks and $9$ green balls to fill them. Thus, the $9$ green balls can be arranged in ${}^{11}{C}_9$ ways.

$\therefore$ Red balls can be arranged in any order.

$\therefore 10!$

Total ways$=10!\times {}^{11}{C}_9$ways.