7 red roses and 3 white roses of different sizes be strung in the form of a garland at random. The probability that no two white roses come together is

7 red roses and 3 white roses of different sizes be strung in the form of a garland at random

$n\left(S\right)=\frac{9!}{2}$

E= Event of arranging flowers so that no two white roses come together 

7 red roses can be arranged in the form of  a garland in 6! ways there are 7 gaps to arrnage 3 white roses in between the red roses in ${ }^7{P}_3$ ways

$n\left(E\right)=6!\times \frac{{ }^7{P}_3}{2}$

$P\left(E\right)=\frac{2}{9!}\times \frac{6!{\times }^7{P}_3}{2}=\frac{6\times 5}{9\times 8}=\frac{5}{12}$