Problem 64
Question
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) \({ }^{30} \mathrm{C}_{8}\)
Step-by-Step Solution
Verified Answer
The correct answer is choice (c): \(\binom{21}{7}\).
1Step 1: 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.
2Step 2: 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.
3Step 3: Calculate remaining marks
After allocating the minimum of 2 marks to each question, we have \(30 - 16 = 14\) marks left to distribute without restrictions.
4Step 4: 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 \(\binom{n+k-1}{k-1}\).
5Step 5: Apply the formula
Using the formula \(\binom{n+k-1}{k-1}\), we substitute \(n = 14\) and \(k = 8\): \[ \text{Number of ways} = \binom{14+8-1}{8-1} = \binom{21}{7} \].
6Step 6: 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): \(\binom{21}{7}\).
Key Concepts
Binomial CoefficientStars and BarsDiscrete Mathematics
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.
Other exercises in this chapter
Problem 62
Let \(\mathrm{T}_{n}\) be the number of all possible triangles formed by joining vertices of an \(n\)-sided regular polygon. If \(\mathrm{T}_{n+1}-\mathrm{T}_{n
View solution Problem 63
On the sides \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) of a \(\triangle \mathrm{ABC}, 3,4,5\) distinct points (excluding vertices \(\mathrm{A}, \mathrm{B}, \mat
View solution Problem 65
A committee of 4 persons is to be formed from 2 ladies, 2 old men and 4 young men such that it includes at least 1 lady, at least 1 old man and at most 2 young
View solution Problem 66
The number of arrangements that can be formed from the letters \(a, b, c, d, e, f\) taken 3 at a time without repetition and each arrangement containing at leas
View solution