I won a trip for four to the Super Bowl.  I can bring three of my friends.  I have 8 friends.  In how many ways can I form my Super Bowl party?
Order does not matter, so it is a combination. Choosing $3$ out of $8$ is $\binom{8}{3}=\boxed{56}.$
56
Level 2
Counting & Probability
Problem: I won a trip for four to the Super Bowl.  I can bring three of my friends.  I have 8 friends.  In how many ways can I form my Super Bowl party?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of ways to choose 3 friends from a group of 8, we need to use the concept of combinations. The order in which we select the friends does not matter, so we apply the combination formula.

The combination formula is given by:

C(n, k) = n! / (k! * (n - k)!)

where n is the total number of items to choose from, k is the number of items to choose, and "!" denotes factorial, which is the product of all positive integers up to that number.

In this scenario, n is 8 (the total number of friends) and k is 3 (the number of friends to choose). 

We can substitute these values into the formula:

C(8, 3) = 8! / (3! * (8 - 3)!)
= 8! / (3! * 5!)

Next, we calculate the factorials:

8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
3! = 3 × 2 × 1 = 6
5! = 5 × 4 × 3 × 2 × 1

Now, we can simplify C(8, 3):

C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1)
= (336) / (6)
= 56

Thus, the total number of ways to form the Super Bowl party by selecting 3 friends from 8 is 56.<|END_OF_LATENT|> The final answer is $56$.