\(\mathrm{C(m,n)} = \frac{\mathrm{m!}}{(\mathrm{m-n})!\mathrm{n!}}\) for nonnegatives integers m and n, m geq n. If \(\mathrm{C(5, 3)}\) = \(\mathrm{C...

1. Translate the problem requirements: We need to understand that \(\mathrm{C(m,n)}\) represents combinations (ways to choose n items from m items), and we're looking for a value \(\mathrm{x≠3}\) where \(\mathrm{C(5,3)}\) equals \(\mathrm{C(5,x)}\).
2. Calculate the reference value: Find \(\mathrm{C(5,3)}\) to establish what value we're matching against.
3. Apply the fundamental combination property: Use the key insight that \(\mathrm{C(m,n) = C(m,m-n)}\), meaning there's always a symmetric pair for any combination.
4. Verify using the symmetry relationship: Confirm that \(\mathrm{C(5,2)}\) equals our reference value \(\mathrm{C(5,3)}\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking in everyday terms.

The formula \(\mathrm{C(m,n)}\) represents combinations - it's asking "how many different ways can I choose n items from a group of m items?" For example, if you have 5 friends and want to choose 3 to invite to dinner, \(\mathrm{C(5,3)}\) tells you how many different groups of 3 friends you could pick.

The problem tells us that \(\mathrm{C(5,3) = C(5,x)}\) where \(\mathrm{x≠3}\). This means we need to find another value x (that's not 3) such that choosing x items from 5 gives us the same number of possibilities as choosing 3 items from 5.

Process Skill: TRANSLATE - Converting the mathematical notation to an understandable real-world scenario

2. Calculate the reference value

Now let's figure out what \(\mathrm{C(5,3)}\) actually equals so we know what we're looking for.

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

Let's calculate step by step:
• \(\mathrm{5! = 5 \times 4 \times 3 \times 2 \times 1 = 120}\)
• \(\mathrm{2! = 2 \times 1 = 2}\)
• \(\mathrm{3! = 3 \times 2 \times 1 = 6}\)

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

This means there are 10 different ways to choose 3 items from 5 items.

3. Apply the fundamental combination property

Here's the key insight that makes this problem much simpler than it first appears:

When you're choosing items to include in a group, you're automatically choosing which items to exclude. If you pick 3 friends out of 5 to invite to dinner, you're simultaneously choosing 2 friends to not invite.

This means that choosing 3 items from 5 is the same as choosing 2 items from 5 - because every way of picking 3 to include corresponds exactly to one way of picking 2 to exclude.

Mathematically, this gives us the property: \(\mathrm{C(m,n) = C(m,m-n)}\)

For our specific case: \(\mathrm{C(5,3) = C(5,5-3) = C(5,2)}\)

Process Skill: INFER - Recognizing the non-obvious symmetry relationship in combinations

4. Verify using the symmetry relationship

Let's confirm that \(\mathrm{C(5,2)}\) indeed equals our reference value of 10:

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

• \(\mathrm{5! = 120}\) (same as before)
• \(\mathrm{3! = 6}\)
• \(\mathrm{2! = 2}\)

So \(\mathrm{C(5,2) = \frac{120}{6 \times 2} = \frac{120}{12} = 10}\) ✓

Perfect! This confirms that \(\mathrm{C(5,3) = C(5,2) = 10}\)

Since we need \(\mathrm{x≠3}\) and \(\mathrm{C(5,x) = C(5,3)}\), we have \(\mathrm{x = 2}\).

Final Answer

The value of x is 2.

This matches answer choice C, which we can verify by checking that it's the only option that makes sense given our symmetry property \(\mathrm{C(5,3) = C(5,5-3) = C(5,2)}\).

Common Faltering Points

Errors while devising the approach

• Missing the constraint \(\mathrm{x≠3}\): Students often focus on solving \(\mathrm{C(5,3) = C(5,x)}\) and find \(\mathrm{x = 3}\) as a solution, but forget that the problem explicitly states \(\mathrm{x≠3}\). This leads them to incorrectly conclude there's no solution or pick the wrong answer.
• Not recognizing the combination symmetry property: Many students attempt to solve this by calculating \(\mathrm{C(5,x)}\) for each answer choice individually, rather than recognizing the fundamental property that \(\mathrm{C(m,n) = C(m,m-n)}\). This makes the problem much more time-consuming and error-prone.
• Misunderstanding what the equation means: Some students might think they need to find multiple values of x or get confused about whether they're looking for one specific value, leading to an unfocused approach to solving the problem.

Errors while executing the approach

• Factorial calculation errors: When computing \(\mathrm{C(5,3) = \frac{5!}{2! \times 3!}}\), students commonly make arithmetic mistakes, especially with \(\mathrm{5! = 120}\) or the denominator calculation \(\mathrm{2! \times 3! = 12}\), leading to an incorrect reference value instead of 10.
• Formula application mistakes: Students may incorrectly apply the combination formula, such as writing \(\mathrm{C(5,3) = \frac{5!}{5! \times 3!}}\) instead of \(\mathrm{C(5,3) = \frac{5!}{2! \times 3!}}\), or confuse the positions of (m-n) and n in the denominator.
• Verification errors: When checking that \(\mathrm{C(5,2)}\) equals the reference value, students might make computational errors or incorrectly conclude that their answer is wrong when it's actually correct, leading them to second-guess their approach.

Errors while selecting the answer

• Selecting \(\mathrm{x = 3}\): Even after correctly finding that both \(\mathrm{x = 2}\) and \(\mathrm{x = 3}\) satisfy the equation, students might forget the constraint \(\mathrm{x≠3}\) and incorrectly select an answer that would correspond to \(\mathrm{x = 3}\) if it were available.
• Confusing the relationship: Students might correctly identify the symmetry but get confused about which value to select, potentially choosing \(\mathrm{5-2 = 3}\) instead of 2, or mixing up the symmetric pair.