Math Problem Statement

A restaurant serves custom-made omelets, where guests select meat, cheese, and vegetables to be added to their omelet. There are 6 vegetables available, and guests may select any 2 vegetables for their omelet. How many different combinations of 2 vegetables are possible?

Solution

To solve this problem, we need to calculate the number of combinations of 2 vegetables that can be selected from the 6 available vegetables.

This is a combination problem, which is calculated using the formula:

C(n,r)=n!r!(nr)!C(n, r) = \frac{n!}{r!(n-r)!}

Here:

  • nn is the total number of items (6 vegetables),
  • rr is the number of items to select (2 vegetables).

Substitute n=6n = 6 and r=2r = 2:

C(6,2)=6!2!(62)!=654!214!C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \cdot 5 \cdot 4!}{2 \cdot 1 \cdot 4!}

The 4!4! cancels out, leaving:

C(6,2)=6521=15C(6, 2) = \frac{6 \cdot 5}{2 \cdot 1} = 15

Answer: There are 15 different combinations of 2 vegetables possible.

Would you like further explanation or examples?


Here are 5 related questions:

  1. How many combinations are possible if guests could select 3 vegetables instead of 2?
  2. What is the formula for determining permutations instead of combinations?
  3. How would the result change if there were 8 vegetables instead of 6?
  4. Can you explain the difference between combinations and permutations?
  5. How many ways can guests select 1 vegetable out of 6?

Tip: Use combinations when the order of items doesn’t matter, and permutations when it does.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Counting Principles

Formulas

C(n, r) = n! / [r! * (n - r)!]

Theorems

Combination formula

Suitable Grade Level

Grades 8-10