How many different numbers can be obtained as the product of exactly 3 different numbers in the set {2,3,5,7,11}

1. Translate the problem requirements: We need to find how many distinct products can be formed by multiplying exactly 3 different numbers chosen from the set \(\{2,3,5,7,11\}\). Since we're looking for products of exactly 3 different numbers, this becomes a combination problem.
2. Recognize the uniqueness property: Since all numbers in the set are distinct prime numbers, any combination of 3 different primes will produce a unique product due to the fundamental theorem of arithmetic.
3. Count the combinations: Determine how many ways we can choose 3 numbers from the 5 available numbers, since each choice gives us exactly one unique product.
4. Calculate using combination formula: Apply \(\mathrm{C(5,3)}\) to find the total number of different products possible.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the question is asking in plain English. We have a set of 5 numbers: \(\{2, 3, 5, 7, 11\}\). We want to pick exactly 3 different numbers from this set and multiply them together. The question asks: how many different products can we get?

For example, if we pick 2, 3, and 5, we get the product: \(2 \times 3 \times 5 = 30\). If we pick 2, 3, and 7, we get: \(2 \times 3 \times 7 = 42\). These are two different products.

Since we need to find products of exactly 3 different numbers, this becomes a problem about choosing groups. We're not concerned with the order we pick the numbers (since multiplication is commutative), just which 3 numbers we select.

Process Skill: TRANSLATE - Converting the product counting problem into a combination selection problem

2. Recognize the uniqueness property

Here's a crucial insight: all the numbers in our set \(\{2, 3, 5, 7, 11\}\) are prime numbers, and they're all different from each other.

Why does this matter? Because of a fundamental property of prime numbers: every number has a unique prime factorization. This means that if we take any 3 different prime numbers and multiply them together, we'll get a product that's different from any other combination of 3 different primes.

For instance:

• \(2 \times 3 \times 5 = 30\)
• \(2 \times 3 \times 7 = 42\)
• \(2 \times 3 \times 11 = 66\)

Each of these products is unique because each uses a different combination of prime factors. There's no way that two different combinations of 3 primes from our set could give us the same product.

Process Skill: INFER - Recognizing that distinct prime combinations guarantee distinct products

3. Count the combinations

Now we know that each different way of choosing 3 numbers from our set of 5 will give us exactly one unique product. So we need to count: how many ways can we choose 3 numbers from 5 numbers?

Let's think about this systematically. We can list out all possible combinations:

Starting with 2: \(\{2,3,5\}\), \(\{2,3,7\}\), \(\{2,3,11\}\), \(\{2,5,7\}\), \(\{2,5,11\}\), \(\{2,7,11\}\)
Starting with 3 (not including 2): \(\{3,5,7\}\), \(\{3,5,11\}\), \(\{3,7,11\}\)
Starting with 5 (not including 2 or 3): \(\{5,7,11\}\)

Let me count these: 6 + 3 + 1 = 10 different combinations.

Each of these combinations gives us a different product, so we have 10 different possible products.

4. Calculate using combination formula

We can verify our counting using the combination formula. When we want to choose 3 items from 5 items (where order doesn't matter), we calculate \(\mathrm{C(5,3)}\).

The combination formula is: \(\mathrm{C(n,r)} = \frac{n!}{r!(n-r)!}\)

So: \(\mathrm{C(5,3)} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \times 2!}\)

Let's calculate this step by step:

• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(2! = 2 \times 1 = 2\)

Therefore: \(\mathrm{C(5,3)} = \frac{120}{6 \times 2} = \frac{120}{12} = 10\)

This confirms our manual counting: there are exactly 10 different ways to choose 3 numbers from our set of 5, which means 10 different products.

Final Answer

We can obtain exactly 10 different numbers as the product of exactly 3 different numbers from the set \(\{2,3,5,7,11\}\).

The answer is B. 10.

This makes sense because we're choosing 3 items from 5 items, and since all the numbers are distinct primes, each combination produces a unique product.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "different numbers" requirement: Students may think they need to find how many different ways to arrange or order the 3 selected numbers, rather than understanding that we're looking for distinct products. This leads them toward permutation thinking (\(\mathrm{P(5,3)} = 60\)) instead of combination thinking.

2. Missing the uniqueness insight: Students may worry that different combinations of 3 numbers could produce the same product and attempt to calculate all possible products manually to check for duplicates. They fail to recognize that since all numbers in the set are distinct primes, each combination automatically produces a unique product.

3. Confusing "exactly 3" with "at most 3": Some students may misinterpret the constraint and try to count products using 1 number, 2 numbers, AND 3 numbers, leading them to calculate \(\mathrm{C(5,1)} + \mathrm{C(5,2)} + \mathrm{C(5,3)} = 5 + 10 + 10 = 25\), which isn't even among the answer choices.

Errors while executing the approach

1. Manual counting errors: When listing out all possible combinations systematically, students often miss combinations or double-count them. For example, they might forget \(\{5,7,11\}\) or accidentally count \(\{2,3,5\}\) and \(\{3,2,5\}\) as different combinations.

2. Combination formula calculation mistakes: Students may make arithmetic errors when computing \(\mathrm{C(5,3)} = \frac{5!}{3! \times 2!}\), such as incorrectly calculating \(5! = 60\) instead of 120, or making errors in the denominator calculation \(3! \times 2! = 6 \times 2 = 12\).

Errors while selecting the answer

1. Verification confusion: Students who correctly calculate \(\mathrm{C(5,3)} = 10\) might second-guess themselves because they expect a larger number of products from 5 numbers, leading them to select a larger answer choice like 15 or 20 instead of trusting their calculation.