Problem 152 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: In an examination consisting of 9 papers, a candidate has to pass in more papers than the numbers of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is 256 . Reason: \({ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots+{ }^{n} C_{n}=2^{n}\)

Step-by-Step Solution

Answer

The calculation supports the assertion as unsuccessful ways are 256.

Step 1: Understand the Assertion

To be successful, the student needs to pass in more than half of the 9 papers. So, the student must pass at least 5 papers. If that's true, success would require passing at least 5 papers. To be unsuccessful, the student must pass in 4 or fewer papers, failing the remaining ones.

Step 2: Determine Number of Unsuccessful Binary Outcomes

For each paper, there are two outcomes: pass or fail. Hence, there are a total of \(2^9\) outcomes possible for 9 papers (each with binary outcomes), which is 512 outcomes.

Step 3: Calculate Unsuccessful Scenarios

Use the binomial coefficient to count the unsuccessful scenarios.- Failing in all 9 papers: \(\binom{9}{9} = 1\)- Failing in 8 papers: \(\binom{9}{8} = 9\)- Failing in 7 papers: \(\binom{9}{7} = 36\)- Failing in 6 papers: \(\binom{9}{6} = 84\)- Failing in 5 papers: \(\binom{9}{5} = 126\)- Failing in 4 papers: \(\binom{9}{4} = 126\)Add these to get: \(1 + 9 + 36 + 84 + 126\), which totals to 256.

Step 4: Verify the Reasoning Provided

The reason \(^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + \ldots + ^{n}C_{n} = 2^{n}\) is true because it accounts for all combinations (where each paper can result in either a pass or fail), hence verifying the calculation of total outcomes, which matches \(2^9 = 512\). Since unsuccessful cases were found to be 256, the number of successful outcomes would be \(512 - 256 = 256\), matching the assertion.

Key Concepts

Combinatorial Mathematics

Combinatorial mathematics is a fascinating field that involves counting and arranging objects. It focuses on the study of finite or countable discrete structures. In our exercise, we explore different combinations of outcomes in an examination scenario where a student attempts nine papers. Each paper can have one of two results: either pass or fail.

In this context, combinatorial mathematics helps by calculating how many specific ways a student can pass or fail a particular number of papers. By using principles from combinatorial mathematics, we determine the number of unsuccessful outcomes by combining different counts of fails and passes. For each of the nine exams, the student has two choices, leading to a more complex calculation of scenarios based on binary decisions.

Understanding combinatorial concepts can be incredibly useful, especially in situations involving choices and arrangements that require careful consideration and calculation.

Binomial Coefficient

The binomial coefficient is a fundamental concept in mathematics, especially in combinatorial calculations. It is denoted as \(\binom{n}{k}\) and represents the number of ways to choose \(k\) items from \(n\) items without regard to the order. In this exercise, binomial coefficients are used to find the number of ways one can fail a specific number of papers out of nine.

For example, if a student wants to know how many ways they can fail exactly four out of nine papers, we use the binomial coefficient \(\binom{9}{4}\), which equals 126. This is computed using the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(!\) denotes factorial, meaning the product of an integer and all the integers below it.

By applying this approach, we calculate various combinations to find out all the possible outcomes in which the number of failed papers is more than half, showing how powerful and essential the use of binomial coefficients in solving combinatorial problems can be.

Probability Theory

Probability theory is a crucial area of mathematics that studies the likelihood and patterns of events occurring. In this exercise, we utilize probability concepts to determine the different scenarios of passing or failing a certain number of papers in an exam.

For nine papers, each with two independent outcomes (pass or fail), there are a total of \(2^9 = 512\) possible outcomes. This total comes from the binary nature of the situation – each paper can result in a pass or fail. Probability theory tells us how these outcomes are distributed; it allows us to calculate how many of these scenarios lead to success or failure.

The exercise specifically examines the unsuccessful outcomes, which occur when a candidate passes fewer papers than they fail. By subtracting the 256 unsuccessful outcomes from the total 512, probability theory helps us understand that there are an equal number of successful outcomes, demonstrating the balance of probabilistic events in binary decision problems. This understanding provides valuable insights into predictable patterns when dealing with randomized events.