Chapter 7: Problem 64
The number of ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question, is: \(\quad\) Online April 22, 2013] (a) \({ }^{30} \mathrm{C}_{7}\) (b) \({ }^{21} C_{8}\) (c) \({ }^{21} C_{7}\) (d) \({ }^{30} \mathrm{C}_{8}\)
Short Answer
Expert verified
The correct answer is choice (c): \(\binom{21}{7}\).
Step by step solution
01
Understand the problem
We need to distribute 30 marks among 8 questions with each question receiving at least 2 marks. The problem asks us to figure out how many ways we can do this distribution.
02
Allocate minimum marks
Since each of the 8 questions must receive at least 2 marks, we begin by giving 2 marks to each question. Therefore, we allocate \(8 \times 2 = 16\) marks initially.
03
Calculate remaining marks
After allocating the minimum of 2 marks to each question, we have \(30 - 16 = 14\) marks left to distribute without restrictions.
04
Formulate as a stars and bars problem
The task now is to distribute 14 additional marks to the 8 questions in any way. This is a typical 'stars and bars' problem, where we find the number of ways to distribute \(n = 14\) indistinguishable items (marks) to \(k = 8\) distinct groups (questions). The number of ways to do this is given by the formula \(\binom{n+k-1}{k-1}\).
05
Apply the formula
Using the formula \(\binom{n+k-1}{k-1}\), we substitute \(n = 14\) and \(k = 8\): \[ \text{Number of ways} = \binom{14+8-1}{8-1} = \binom{21}{7} \].
06
Find the correct answer choice
The answer from the calculation is \(\binom{21}{7}\). Check the provided answer choices: - Choice (a): \(\binom{30}{7}\) - Choice (b): \(\binom{21}{8}\) - Choice (c): \(\binom{21}{7}\) - Choice (d): \(\binom{30}{8}\) Thus, the correct answer is choice (c): \(\binom{21}{7}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Coefficient
The concept of a binomial coefficient, often denoted as \( \binom{n}{k} \), is foundational in combinatorics. It represents the number of ways to choose \( k \) items from a total of \( n \) items without regard to the order of selection.
To calculate a binomial coefficient, we use the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \).
To calculate a binomial coefficient, we use the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \).
- For example, \( \binom{5}{2} \) equates to finding the number of ways to choose 2 items from 5, which equals 10.
- The binomial coefficient is symmetric, meaning \( \binom{n}{k} = \binom{n}{n-k} \).
Stars and Bars
The stars and bars method is a powerful technique in combinatorics for distributing indistinguishable items into distinct groups.
In our problem, we distributed 14 remaining marks (stars) among 8 questions (buckets or groups). Here's how it works:
Applying this to our scenario—distributing 14 marks to 8 questions—the formula turns into \( \binom{14+8-1}{8-1} = \binom{21}{7} \). This approach simplifies complex distribution tasks by converting them into a simple problem of choosing positions for dividers (bars) among the items (stars).
In our problem, we distributed 14 remaining marks (stars) among 8 questions (buckets or groups). Here's how it works:
- The stars represent what you are distributing—in our case, marks.
- The bars serve as dividers between different groups, here the questions.
Applying this to our scenario—distributing 14 marks to 8 questions—the formula turns into \( \binom{14+8-1}{8-1} = \binom{21}{7} \). This approach simplifies complex distribution tasks by converting them into a simple problem of choosing positions for dividers (bars) among the items (stars).
Discrete Mathematics
Discrete mathematics is a field of mathematics that deals with countable, distinct elements. It includes a wide array of topics like combinatorics, graph theory, and logic.
In this exercise, we are primarily concerned with combinatorics, a critical topic within discrete mathematics, which deals with counting, arrangement, and combination of elements. Some essential areas of discrete mathematics include:
In this exercise, we are primarily concerned with combinatorics, a critical topic within discrete mathematics, which deals with counting, arrangement, and combination of elements. Some essential areas of discrete mathematics include:
- Understanding structures that are fundamentally discrete rather than continuous.
- Utilizing discrete structures for the analysis of problems in computer science, such as algorithm efficiency and data organization.