There are 10 red and 5 yellow roses of differ | vedclass

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(A) Total roses: $10$ red and $5$ yellow. Total flowers = $15$.For garlands,the number of arrangements of $n$ distinct items is $\frac{(n-1)!}{2}$.$x$

Vedclass · Accepted Answer

$\frac{9!}{5!}-5!$

Vedclass · Answer

(A) Total roses: $10$ red and $5$ yellow. Total flowers = $15$.For garlands,the number of arrangements of $n$ distinct items is $\frac{(n-1)!}{2}$.$x$: No two yellow roses come together. First,arrange $10$ red roses in a circle in $\frac{(10-1)!}{2} = \frac{9!}{2}$ ways. There are $10$ gaps created between them. We arrange $5$ yellow roses in these $10$ gaps in $P(10, 5)$ ways.$x = \frac{9!}{2} 	imes P(10, 5) = \frac{9!}{2} 	imes \frac{10!}{5!}$.$y$: All red roses come together. Treat $10$ red roses as $1$ unit. Now we have $1$ unit of red roses and $5$ individual yellow roses,total $6$ items. Arrange these in a circle in $\frac{(6-1)!}{2} = \frac{5!}{2}$ ways. The $10$ red roses can be arranged among themselves in $10!$ ways.$y = \frac{5!}{2} 	imes 10!$.Now,$\frac{2(x-y)}{10!} = \frac{2}{10!} \left( \frac{9! 	imes 10!}{2 	imes 5!} - \frac{5! 	imes 10!}{2} ight) = \frac{9!}{5!} - 5!$.

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