Question is:
in a window display at a flower shop, there are 3 spots for 1 plant each. to fill these 3 spots, adam has 7 plants to select from, each of a different type. Selecting from the 7 plants, adam can make how many display arrangements with 1 plant in each spot? (Note: the positions of the unselected plants do not matter.)
If the order does not matter, shouldn't this be a combination? nCr would be 35. the "correct" answer is J (210), but if I understand correctly, that only occurs with a permutation of 765, which is calculating when order DOES matter. I dont understand why this would be 210 if order clearly does not matter. Any help is appreciated thanks!
![[1]: https://i.stack.imgur.com/aG8wf.png](https://i.stack.imgur.com/zZqaP.png)
Best Answer
There are $\binom73$ ways to choose $3$ of the plants to be displayed, but there are then $3!$ ways to arrange those $3$ plants in the $3$ display spots. Thus, there are altogether
$$\binom73\cdot3!=35\cdot6=210$$
possible displays.
Choosing the $3$ plants is choosing a combination of plants, but the $3$ spots in the window have distinct identities: their positions are fixed. Thus, if the plants are $A,B$, and $C$, it makes a difference whether we arrange them in the $3$ spots in the order $ABC$, the order $BAC$, or one of the other four orders.