You have six different flavors of ice cream and want to choose 3 different scoops. Since the order in which you choose the scoops doesn't matter, this is a combination problem.

The formula for combinations is:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

where:

• $n$ is the total number of items (6 flavors),
• $r$ is the number of selections to be made (3 scoops).

Substituting the values into the formula:

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}$

Simplifying:

$= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20$

So, there are 20 different combinations of 3 scoops from 6 different flavors.

Would you like more details on combinations or have any other questions?

Here are 5 related questions:

1. How would the number of combinations change if you could repeat flavors?
2. What if you had to choose 2 scoops instead of 3?
3. How does the combination formula change for larger sets?
4. What if the scoops were selected in a specific order?
5. How does this compare to choosing 3 scoops from 7 flavors?

Tip: In combination problems, order does not matter; in permutation problems, order matters.