Question: there are 2 sections a b c with 5 questions each such that one question from each section is compuls...

Question

there are 2 sections a b c with 5 questions each such that one question from each section is compulsory and a total of 5q can be answered find number of ways

Answer 1

Solution

The problem asks us to find the number of ways to answer 5 questions from three sections A, B, C, each containing 5 questions, with the conditions that one question from each section is compulsory, and a total of 5 questions must be answered.

Let $n_A$ , $n_B$ , and $n_C$ be the number of questions chosen from section A, B, and C respectively.

The given conditions can be stated as:

Compulsory Question: At least one question must be chosen from each section. So, $n_A \ge 1$ , $n_B \ge 1$ , and $n_C \ge 1$ .
Total Questions: A total of 5 questions must be answered. So, $n_A + n_B + n_C = 5$ .
Maximum Questions per Section: Each section has 5 questions. So, $n_A \le 5$ , $n_B \le 5$ , and $n_C \le 5$ .

We need to find all possible combinations of non-negative integers $(n_A, n_B, n_C)$ that satisfy these conditions. Since $n_A+n_B+n_C=5$ and each $n_i \ge 1$ , the maximum value any $n_i$ can take is $5 - 1 - 1 = 3$ . This means $n_i \le 3$ , which automatically satisfies the $n_i \le 5$ condition.

Let's list the possible sets of $(n_A, n_B, n_C)$ that sum to 5, with each component being at least 1:

Case 1: (1, 1, 3) This means we choose 1 question from section A, 1 from section B, and 3 from section C. The number of ways to choose questions for this specific distribution is: $\binom{5}{1}$ (for section A) $\times \binom{5}{1}$ (for section B) $\times \binom{5}{3}$ (for section C) $= 5 \times 5 \times 10 = 250$ ways.

Since the sections are distinct, the distribution (1, 1, 3) can be permuted in three ways:

(1, 1, 3): 1 from A, 1 from B, 3 from C. Ways = 250.
(1, 3, 1): 1 from A, 3 from B, 1 from C. Ways = $\binom{5}{1} \times \binom{5}{3} \times \binom{5}{1} = 5 \times 10 \times 5 = 250$ .
(3, 1, 1): 3 from A, 1 from B, 1 from C. Ways = $\binom{5}{3} \times \binom{5}{1} \times \binom{5}{1} = 10 \times 5 \times 5 = 250$ . Total ways for this type of distribution = $3 \times 250 = 750$ .

Case 2: (1, 2, 2) This means we choose 1 question from section A, 2 from section B, and 2 from section C. The number of ways to choose questions for this specific distribution is: $\binom{5}{1}$ (for section A) $\times \binom{5}{2}$ (for section B) $\times \binom{5}{2}$ (for section C) $= 5 \times 10 \times 10 = 500$ ways.

This distribution (1, 2, 2) can also be permuted in three ways:

(1, 2, 2): 1 from A, 2 from B, 2 from C. Ways = 500.
(2, 1, 2): 2 from A, 1 from B, 2 from C. Ways = $\binom{5}{2} \times \binom{5}{1} \times \binom{5}{2} = 10 \times 5 \times 10 = 500$ .
(2, 2, 1): 2 from A, 2 from B, 1 from C. Ways = $\binom{5}{2} \times \binom{5}{2} \times \binom{5}{1} = 10 \times 10 \times 5 = 500$ . Total ways for this type of distribution = $3 \times 500 = 1500$ .

Total Number of Ways: The total number of ways to answer the questions is the sum of ways from all possible distributions: Total Ways = (Total ways from Case 1) + (Total ways from Case 2) Total Ways = $750 + 1500 = 2250$ .

The interpretation of "one question from each section is compulsory" means that the final selection must contain at least one question from each section. The method used above correctly accounts for this by setting $n_A, n_B, n_C \ge 1$ .