Find the number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland. In how many of them
i) All yellow roses together
ii) No two yellow roses together

No. of red roses = 6
No. of yellow roses = 3
Total no. of roses = 6 + 3 = 9
$\therefore$ The number of ways of arranging the 9 roses
in a garland $=\frac{1}{2}(9-1)!=\frac{1}{2}\times 8!$ ways

(i) All the yellow roses are together:-
Treat all the 3 yellow roses as one unit then we have 6 red roses + 1 unit = 7 units They can be arranged in a garland in$\frac{1}{2}(7-1)!=\frac{1}{2}\times 6!$ ways

Now the 3 yellow roses can be arranged among themselves in 3! ways

$\therefore$Req. no. of arrangements$=\frac{1}{2}\times 6!\times 3!$

(ii) No two yellow roses are together:-
First we arrange the 6 red roses in a garland in $\frac{1}{2}(6-1)!$  ways then we can find 6 gaps among them the 3 yellow roses can be arranged in these 6 gaps in ${ }^6{P}_3$ways

$\therefore$Req. no. of arrangements $=\frac{1}{2}\times (6-1)!\times {}^6{P}_3$

$=\frac{1}{2}\times 5!\times {}^6{P}_3$