Calculate the number of possible ice cream flavor combinations given 30 total flavors and 3 scoops to choose | Step-by-Step Solution
Problem
Robert goes to an ice cream shop with 30 flavors and can get 3 different scoops. How many different combinations can he get, both with and without flavor repetition?
🎯 What You'll Learn
- Understand combination calculation methods
- Learn to distinguish between combinations with and without repetition
- Apply mathematical reasoning to real-world scenarios
Prerequisites: Basic multiplication, Understanding of factorials, Fundamental counting principles
💡 Quick Summary
Great question! This is a classic combinatorics problem that deals with choosing items from a set, and the key insight is recognizing whether we're allowing repeated flavors or not. Think about this: does it matter if Robert gets his scoops in the order vanilla-chocolate-strawberry versus chocolate-vanilla-strawberry, or is that considered the same combination? Also, can Robert choose the same flavor multiple times (like three scoops of vanilla), or must all three scoops be different flavors? Once you figure out these two important details, you'll want to look into combination formulas - there's one for when repetition isn't allowed and a slightly different one (involving "combinations with replacement") for when it is allowed. Try sketching out a few small examples first, maybe with just 4-5 flavors, to see the pattern before jumping into the full calculation with 30 flavors!
Step-by-Step Explanation
What We're Solving:
We need to find how many different 3-scoop ice cream combinations Robert can make from 30 flavors in TWO scenarios: when he can repeat flavors and when he cannot.The Approach:
The key insight here is recognizing that we have TWO different situations that require different mathematical approaches:- Without repetition: This is a combination problem (order doesn't matter, can't repeat)
- With repetition: This is a combination with replacement problem (order doesn't matter, can repeat)
Step-by-Step Solution:
Case 1: WITHOUT flavor repetition (all 3 scoops must be different flavors)
- We're choosing 3 different flavors from 30 available flavors
- Use the combination formula: C(n,r) = n! / (r!(n-r)!)
- Here: C(30,3) = 30! / (3!(30-3)!) = 30! / (3! × 27!)
- This simplifies to: (30 × 29 × 28) / (3 × 2 × 1)
- Calculate: (24,360) / (6) = 4,060 combinations
- Robert could get 3 vanilla scoops, or 2 chocolate + 1 mint, etc.
- This uses combinations with replacement: C(n+r-1, r)
- Here: C(30+3-1, 3) = C(32, 3)
- Calculate: C(32,3) = (32 × 31 × 30) / (3 × 2 × 1)
- This gives us: (29,760) / (6) = 4,960 combinations
The Answer:
- Without repetition: 4,060 different combinations
- With repetition: 4,960 different combinations
Memory Tip:
Remember "Stars and Bars" for repetition problems! When you can repeat items, imagine you have extra "slots" to work with. That's why we use C(n+r-1, r) instead of just C(n,r). The extra slots represent the possibility of picking the same flavor multiple times!You've got this! Combinatorics gets easier once you can identify whether order matters and whether repetition is allowed. 🌟
⚠️ Common Mistakes to Avoid
- Overcalculating by trying to list all possibilities
- Confusing permutations with combinations
- Not accounting for repetition rules
This explanation was generated by AI. While we work hard to be accurate, mistakes can happen! Always double-check important answers with your teacher or textbook.
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📷 Problem detected:
Solve: 2x + 5 = 13
Step 1:
Subtract 5 from both sides...
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