Mastering Combinations: Fun Quiz & Solutions!

Solve the following questions. These questions focus on different aspects of combinations, including calculating combinations, understanding their applications, and recognizing related concepts. I'll assume you're familiar with the basic formula for combinations: nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items chosen.

 

1. Question: How many ways can you choose a committee of 3 people from a group of 10?

 

Answer: 120

 

Solution: This is a combination problem because the order in which the committee members are chosen doesn't matter. The solution is 10C3 = 10! / (3!(10-3)!) = 120

 

2. Question: A pizza place offers 5 toppings. How many different 2-topping pizzas can you order?

 

Answer: 10

 

Solution: Again, the order of the toppings doesn't matter (pepperoni and mushrooms is the same as mushrooms and pepperoni). The calculation is 5C2 = 5! / (2!(5-2)!) = 10

 

3. Question: A deck of cards has 52 cards. How many 5-card hands are possible?

 

Answer: 2,598,960

 

Solution: 52C5 = 52! / (5!(52-5)!) = 2,598,960

 

4. Question: You have 7 books. How many ways can you arrange 3 of them on a shelf?

 

Answer: This is not a combination problem. This is a permutation problem because the order matters. The answer is 7P3 = 210.

 

Solution: Permutations consider order, while combinations do not. The formula for permutations is nPr = n! / (n-r)!.

 

5. Question: A lottery has 49 numbers. You choose 6. How many different combinations are possible?

 

Answer: 13,983,816

 

Solution: 49C6 = 49! / (6!(49-6)!) = 13,983,816

 

6. Question: What is the difference between a combination and a permutation?

 

Answer: Combinations are selections where order doesn't matter; permutations are selections where order does matter.

 

Solution: Understanding this fundamental difference is key to solving combination and permutation problems correctly.

 

7. Question: A teacher has 20 students. They need to choose 5 students to represent the class at a competition. How many different teams are possible?

 

Answer: 15,504

 

Solution: 20C5 = 20! / (5!(20-5)!) = 15,504

 

8. Question: You have 12 colored pencils. How many ways can you choose 4 to use for a drawing?

 

Answer: 495

 

Solution: 12C4 = 12! / (4!(12-4)!) = 495

 

9. Question: A basketball team has 15 players. How many ways can the coach choose a starting lineup of 5 players? (Assume all positions are equivalent.)

 

Answer: 3003

 

Solution: 15C5 = 15! / (5!(15-5)!) = 3003

 

10. Question: Explain why the number of combinations of choosing 2 items from a set of 5 is the same as the number of combinations of choosing 3 items from that same set of 5.

 

Answer: Because choosing 2 items implicitly leaves 3 items unchosen. The number of ways to choose 2 is the same as the number of ways to choose which 3 to not choose.

 

Solution: This highlights the symmetry inherent in combination calculations. nCr = nC(n-r)

 

These questions provide a range of difficulty and application scenarios for combinations. Remember to carefully identify whether a problem requires combinations or permutations before attempting to solve it.