Find the number of ways in which 6 red roses and 3 white roses (all roses different) can form a garland such that all the white roses come together
Answer
646.2k+ views
Hint:
Here in this problem we use the formula for circular permutation.
Complete step by step solution:
There are 6 red roses and 3 white roses (all roses different) which need to be formed into a garland.
The formula for circular permutation of n distinct objects into an arrangement which can be flipped is \[\dfrac{{\left( {n - 1} \right)!}}{2}\]. So considering the three white roses as one element, and the six red roses, there are seven distinct elements.
Hence the number of permutations is \[\dfrac{{\left( {7 - 1} \right)!}}{2}\].
Again, the white roses also arrange amongst themselves, so, the total number of arrangements is
\[\begin{array}{l}
\dfrac{{\left( {7 - 1} \right)!}}{2}\left( {3!} \right)\\
= \dfrac{{6!}}{2}\left( {3!} \right)\\
= 6! \times 3\\
= 2160
\end{array}\]
Note:
In these types of questions it is important to understand that it is a case of circular permutation and not linear permutation.
Here in this problem we use the formula for circular permutation.
Complete step by step solution:
There are 6 red roses and 3 white roses (all roses different) which need to be formed into a garland.
The formula for circular permutation of n distinct objects into an arrangement which can be flipped is \[\dfrac{{\left( {n - 1} \right)!}}{2}\]. So considering the three white roses as one element, and the six red roses, there are seven distinct elements.
Hence the number of permutations is \[\dfrac{{\left( {7 - 1} \right)!}}{2}\].
Again, the white roses also arrange amongst themselves, so, the total number of arrangements is
\[\begin{array}{l}
\dfrac{{\left( {7 - 1} \right)!}}{2}\left( {3!} \right)\\
= \dfrac{{6!}}{2}\left( {3!} \right)\\
= 6! \times 3\\
= 2160
\end{array}\]
Note:
In these types of questions it is important to understand that it is a case of circular permutation and not linear permutation.
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