Problem 153 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is (A) 140 (B) 196 (C) 280 (D) 346

Step-by-Step Solution

Answer

The student has 196 ways to choose the questions, so (B) 196 is the answer.

Step 1: Define the problem

A student has to choose 10 questions out of a total of 13, where these questions are divided into two groups: 5 in the first group and 8 in the second group. The student has to choose at least 4 questions from the first group of 5 questions.

Step 2: Analyze constraints and subsets

Since the student must choose at least 4 from the first 5 questions, we consider two cases: one where the student selects exactly 4 questions from the first 5, and another where the student selects all 5 questions from the first group. Depending on these cases, the student then selects the remaining questions from the second group.

Step 3: Calculate possibilities for Case 1

**Case 1**: The student selects 4 questions from the first 5. We can calculate the number of ways to choose 4 questions out of 5 using combinations, denoted as \( \binom{5}{4} \).Calculate: \( \binom{5}{4} = 5 \).Now, the student needs to select 6 more questions from the remaining 8 questions in the second group, which is calculated using \( \binom{8}{6} \).Calculate: \( \binom{8}{6} = 28 \).Thus, the total number of ways for Case 1 is \( 5 \times 28 = 140 \).

Step 4: Calculate possibilities for Case 2

**Case 2**: The student selects all 5 questions from the first 5. This can be done in \( \binom{5}{5} = 1 \) way.Now, the student needs to select 5 more questions from the remaining 8 questions in the second group, which is calculated using \( \binom{8}{5} \).Calculate: \( \binom{8}{5} = 56 \).Thus, the total number of ways for Case 2 is \( 1 \times 56 = 56 \).

Step 5: Combine the results

Add the results from both cases to get the total number of ways the student can choose the questions.Total ways = \( 140 + 56 = 196 \).

Key Concepts

Combinations

When discussing combinations, we refer to the process of selecting items from a larger pool where the order of selection does not matter. This concept is primarily represented by the binomial coefficient, often denoted as \( \binom{n}{r} \). Here, \( n \) is the total number of items to choose from, and \( r \) is the number of items to select. The formula to calculate combinations is:

• \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

In the exercise, we see combinations used to determine how many ways a student can select questions from a set of examination problems. For instance, calculating \( \binom{5}{4} \) means choosing 4 questions from the available 5, and this concept is also applied to find how many more questions to choose from the remaining options in the second set.

Binomial Theorem

The binomial theorem is a fundamental concept connecting algebra and combinatorics. It reveals the expanded form of a binomial raised to any positive integer power. For example, the expression \((a + b)^n\) can be expanded using the theorem:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)

Each term in the expansion is formed by a combination, which highlights why the binomial coefficient is central in combinatorics. While the binomial theorem doesn't directly solve the exercise, understanding its connection to combinations helps clarify why such coefficients are used to count possibilities in scenarios like the exam questions.

Permutations

Permutations differ from combinations in that the order of selection matters. When calculating permutations, we're interested in the different ways items can be arranged. The permutation formula is:

• \( nPr = \frac{n!}{(n-r)!} \)

Although permutations are not directly applied in the given problem, it's important to understand the difference. If the student needed to arrange the selected questions in a specific order, permutations would be required. Recognizing when to use permutations versus combinations is crucial in solving problems correctly and efficiently in combinatorics.