There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections...

1. Translate the problem requirements: We need selections of exactly 5 books from 10 total books (4 paperbacks + 6 hardbacks) where each selection must include at least one paperback AND at least one hardback
2. Identify the complementary counting approach: Rather than directly counting valid selections, calculate total possible selections and subtract the invalid ones (selections with only paperbacks or only hardbacks)
3. Calculate total unrestricted selections: Find the number of ways to choose any 5 books from the 10 available books
4. Calculate and subtract invalid selections: Find selections that violate our requirements (all paperbacks only, or all hardbacks only) and remove them from the total

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down exactly what we're looking for in plain English. We have a shelf with 10 books total: 4 paperbacks and 6 hardbacks. We want to select exactly 5 books, but not just any 5 books - our selection must include at least one paperback AND at least one hardback.

Think of it this way: if you grabbed 5 books randomly, some selections would be "good" (having both types of books) and some would be "bad" (having only one type). We need to count only the "good" selections.

Process Skill: TRANSLATE - Converting the constraint "at least one of each type" into a clear mathematical requirement

2. Identify the complementary counting approach

Here's the key insight: instead of trying to directly count all the "good" selections (which would be complicated), we can count ALL possible selections and then subtract the "bad" ones.

What makes a selection "bad"? It's when we pick:

- Only paperbacks (no hardbacks), OR
- Only hardbacks (no paperbacks)

So our approach is: Good selections = Total selections - Bad selections

This is much easier than trying to count all the different combinations of paperbacks and hardbacks directly!

Process Skill: CONSIDER ALL CASES - Recognizing that complementary counting avoids complex case analysis

3. Calculate total unrestricted selections

First, let's find how many ways we can choose any 5 books from the 10 available books, with no restrictions.

This is asking: "In how many ways can I choose 5 items from 10 items?"

Using the combination formula: \(\mathrm{C(10,5)} = \frac{10!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}\)

Let's calculate step by step:

- Numerator: \(10 \times 9 \times 8 \times 7 \times 6 = 30,240\)
- Denominator: \(5 \times 4 \times 3 \times 2 \times 1 = 120\)
- Total: \(30,240 \div 120 = 252\)

So there are 252 ways to choose any 5 books from 10 books.

4. Calculate and subtract invalid selections

Now we need to subtract the "bad" selections - those that have only one type of book.

Case 1: Selections with only paperbacks
We have 4 paperbacks total, and we want to choose 5 books. Since we only have 4 paperbacks, it's impossible to choose 5 books using only paperbacks. So this contributes 0 invalid selections.

Case 2: Selections with only hardbacks
We have 6 hardbacks total, and we want to choose 5 books using only hardbacks.
This is: \(\mathrm{C(6,5)} = \frac{6!}{5! \times 1!} = 6\)

So there are 6 ways to choose 5 books using only hardbacks.

Final calculation:
Valid selections = Total selections - Invalid selections
Valid selections = \(252 - 0 - 6 = 246\)

5. Final Answer

The number of possible selections of 5 books that contain at least one paperback and at least one hardback is 246.

This matches answer choice D) 246.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "at least one of each type" constraint
Students often misread "at least one paperback AND at least one hardback" as "exactly one paperback AND exactly one hardback" or think they need to count specific combinations like "1 paperback + 4 hardbacks" separately. This leads them to attempt complex direct counting instead of recognizing the complementary counting opportunity.

2. Attempting direct counting instead of complementary counting
Many students try to directly count valid selections by considering cases like "1 paperback + 4 hardbacks," "2 paperbacks + 3 hardbacks," etc. This approach is much more complex and error-prone. They miss that it's easier to count what we DON'T want (all paperbacks OR all hardbacks) and subtract from the total.

3. Forgetting to check feasibility of invalid cases
Students may set up the complementary counting approach correctly but fail to recognize that selecting 5 books from only 4 paperbacks is impossible. They might attempt to calculate \(\mathrm{C(4,5)}\) without realizing this equals zero, leading to incorrect subtraction.

Errors while executing the approach

1. Combination formula calculation errors
Students frequently make arithmetic mistakes when calculating \(\mathrm{C(10,5)}\), especially in the step-by-step multiplication and division. Common errors include miscalculating \(10\times9\times8\times7\times6 = 30,240\) or incorrectly dividing by \(5! = 120\), leading to wrong values for total selections.

2. Incorrectly handling impossible combinations
When calculating \(\mathrm{C(4,5)}\) for "all paperbacks" selections, students might not recognize this is impossible and either try to force a calculation or assign it a non-zero value, leading to incorrect subtraction in the final step.

Errors while selecting the answer

1. Arithmetic error in final subtraction
Even with correct intermediate calculations (252 total selections and 6 invalid "all hardback" selections), students may make simple arithmetic errors in the final step: \(252 - 6 = 246\), potentially getting 258 or other incorrect values due to careless calculation.