In the local ice cream shop, there are 10 different flavors. How many
different two-scoop cones are there? Here, the order does not matter.
A cone with a vanilla scoop on top of a chocolate scoop is considered
the same as a cone with a chocolate scoop on top of a vanilla scoop.
Here's my approach so far:
I'm choosing 2 flavors from 10, so I use ${10\choose 2} = 45$ different two-scoop cones.
Now suppose that the order of the scoops does matter. How many
different possible two-scoop cones are there?
This time, there are even more cones that have been excluded previously, and I use permutation:
$\frac{10!}{8!} = 90$ different two-scoop cones.
My issue is that I was told the answers were 100 and 55, respectively.
The only way I could assume 100 is correct for the first one is if you could repeat flavors. That way the number of different cones would be $10*10 = 100$. But if that's the case, then how would I calculate 55 for the second problem? My method seems to fall apart.
Any guidance would be appreciated, thank you.
Best Answer
You've been told the answers the wrong way round. The first case has $45$ cones with two different scoops, plus $10$ where the flavours are the same.
The second case has $10\times10=100$ cones.