Question

The number of ways in which $8$ red roses and $5$ white roses of different sizes can be made out to form a garland so that no two white roses come together is

• A. $\frac{8!}{2!}{.}^8{P}_5$
• B. $\frac{7!}{2!}{.}^8{P}_5$
• C. $\frac{7!}{2!}{.}^9{P}_5$
• D. $7!{.}^4{P}_3$

Answer 1

Answer: (B)

$\frac{7!}{2!}{.}^8{P}_5$

As no two white roses are to come together , it is taken as arranging white roses between red roses.
$no$ of way of arranging in a garland $=\frac{(n-1)!}{2}$
$\therefore no$ of ways of arranging red roses$=\frac{(n-1)!}{2}$$=\frac{7!}{2}$

Number of places in between red roses is 8
$no$ of ways of placing 5 roses in them is ${}^8{P}_5$
$\therefore$ Total $no$ of ways $=\frac{7!}{2}{\times }^8{P}_5$