Suppose I have 3 types of flowers and I want to have 2 flowers.
Shouldn't the answer be $(3 * 3)/2!$ as I have 3 ways to select 1st flower, 3 for 3nd flower and as I don't care about order, I divide by $2!$.
But the answer here will comee out to be 4.5 which is not possible.
Where is my reasoning wrong?
Thank you.
The problem is that there is only one way of choosing the first type of flower $f_1$ twice, and similarly for $f_2$ and $f_3$. On the other hand, you can choose different flowers, e.g. flower 1 and flower 2, in two distinct ways, namely $(f_1,f_2)$ and $(f_2,f_1)$. You could count the number $N$ of different combinations as follows instead: Choosing the same flower twice can be done in one way, and choosing any two different flowers can be done in two ways (which you want to consider as the same way, hence the factor $\tfrac12$ below), so
\begin{align*} N = 3\cdot 1 + 2\cdot{{3}\choose{2}}\cdot\frac12 = 6. \end{align*}