Question

In how many ways can $4$ red, $3$ yellow and $2$ green discs be arranged in a row. If the discs of the same colour are indistinguishable?

Answer 1

Total number of discs $=4\text{red}+3\text{yellow}+2\text{green}$
$\therefore n=9$
$\because$ Color is repeating so,
$n=9,P_1=4\text{(red)},P_2=3\text{(yellow) and}{P}_3=2\text{(green)}$
$\therefore$ Total number of arrangement$=\frac{n!}{P_1!P_2!P_3!}$
$=\frac{9!}{4!3!2!}$
$=\frac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 3\times 2\times 1\times 2\times 1}$
$=1260$

Hence, total number of arrangements is $1260$.