Maths Q16 – AP-EAMCET 2026

Home Maths Permutations and CombinationsQuestion #310

A candidate is required to answer seven out of 12 questions which are divided into two parts, each containing 6 questions. If he is not permitted to answer more than 5 questions from each part, the different number of ways he can choose the 7 questions is

Clue 1 of 3: Problem Decomposition

Break down the problem into smaller, manageable parts by first identifying the total number of questions to be answered and the constraints on choosing questions from each section.

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Step 1: Analyze the Constraints and Identify Variables✦ Active

The candidate must answer 7 questions out of 12. The 12 questions are divided into two parts, Part 1 and Part 2, each containing 6 questions. Let $x$ be the number of questions answered from Part 1 and $y$ be the number of questions answered from Part 2. We have the following conditions:

x + y = 7

The candidate is not permitted to answer more than 5 questions from each part, meaning:

x \leq 5 and y \leq 5

Also, since there are 6 questions in each part, $x$ and $y$ must also satisfy $x \leq 6$ and $y \leq 6$ . However, $x \leq 5$ and $y \leq 5$ are stricter conditions, so we primarily use these.

💡 Teacher's Secret Hint

Always start by clearly listing all given conditions and what needs to be found. This helps prevent oversight in complex problems.

Step 2: Determine All Valid Combinations for

(x, y)

○ Expand

We need to find pairs of non-negative integers $(x, y)$ such that $x + y = 7$ , $x \leq 5$ , and $y \leq 5$ . Let's list the possibilities for $x$ and corresponding $y$ values:

If $x = 1$ , then $y = 6$ . This violates $y \leq 5$ . (Invalid)

If $x = 2$ , then $y = 5$ . This satisfies all conditions ( $2 \leq 5$ , $5 \leq 5$ ). (Valid)

If $x = 3$ , then $y = 4$ . This satisfies all conditions ( $3 \leq 5$ , $4 \leq 5$ ). (Valid)

If $x = 4$ , then $y = 3$ . This satisfies all conditions ( $4 \leq 5$ , $3 \leq 5$ ). (Valid)

If $x = 5$ , then $y = 2$ . This satisfies all conditions ( $5 \leq 5$ , $2 \leq 5$ ). (Valid)

If $x = 6$ , then $y = 1$ . This violates $x \leq 5$ . (Invalid)

So, the valid combinations for $(x, y)$ are $(2, 5), (3, 4), (4, 3), (5, 2)$ .

Step 3: Calculate Ways for Each Valid Combination○ Expand

We use the combination formula $(\binom{n}{k}) = \frac{n!}{k! (n - k)!}$ to find the number of ways to choose questions from each part. Each part has 6 questions.

Case 1: $x = 2$ from Part 1 and $y = 5$ from Part 2.

(\binom{6}{2}) \times (\binom{6}{5}) = \frac{6 \times 5}{2 \times 1} \times \frac{6}{1} = 15 \times 6 = 90

Case 2: $x = 3$ from Part 1 and $y = 4$ from Part 2.

(\binom{6}{3}) \times (\binom{6}{4}) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{6 \times 5}{2 \times 1} = 20 \times 15 = 300

Case 3: $x = 4$ from Part 1 and $y = 3$ from Part 2.

(\binom{6}{4}) \times (\binom{6}{3}) = \frac{6 \times 5}{2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 15 \times 20 = 300

Case 4: $x = 5$ from Part 1 and $y = 2$ from Part 2.

(\binom{6}{5}) \times (\binom{6}{2}) = \frac{6}{1} \times \frac{6 \times 5}{2 \times 1} = 6 \times 15 = 90

💡 Teacher's Secret Hint

Remember that $(\binom{n}{k}) = (\binom{n}{n - k})$ , which can simplify calculations. For example, $(\binom{6}{4}) = (\binom{6}{2})$ and $(\binom{6}{5}) = (\binom{6}{1})$ .

Step 4: Calculate Total Number of Ways○ Expand

To find the total number of ways the candidate can choose 7 questions, we sum the ways from all valid cases:

Total ways = 90 + 300 + 300 + 90 = 780

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Maths Question 16 – AP-EAMCET 2026

Clue 1 of 3: Problem Decomposition

🧠 Full Solution Path