Problem 110 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of ways in which 30 marks can be alloted to 8 questions if each question carries at least 2 marks, is (A) 115280 (B) 117280 (C) 116280 (D) None of these

Step-by-Step Solution

Answer

There are 116280 ways to distribute the marks, which matches option (C).

Step 1: Understand the Problem

We need to distribute 30 marks among 8 questions, where each question must receive at least 2 marks. This is a combinatorial distribution problem with constraints.

Step 2: Adjust for Minimum Marks per Question

Since each question must receive at least 2 marks, assign 2 marks to each question first. This uses up a total of 16 marks (since 2 marks/question * 8 questions = 16 marks).

Step 3: Calculate Remaining Marks

After allocating the minimum required marks, we have 30 - 16 = 14 marks remaining, which need to be distributed among the 8 questions with no restrictions.

Step 4: Use Stars and Bars Method

To distribute 14 marks among 8 questions (unrestricted distribution), apply the stars and bars method. The formula for distributing n identical items into k groups is \[ \binom{n + k - 1}{k - 1} \]Here, n = 14 (remaining marks) and k = 8 (questions), so we calculate \[ \binom{14 + 8 - 1}{8 - 1} = \binom{21}{7} \]

Step 5: Compute the Binomial Coefficient

Calculate the value of \( \binom{21}{7} \): - Split the calculation to simplify: \[ \binom{21}{7} = \frac{21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \]- Compute: \( = \frac{2030937600}{5040} = 116280 \)

Step 6: Conclusion

Thus, the number of ways to distribute 30 marks to 8 questions, each with at least 2 marks, is 116280.

Key Concepts

Stars and Bars Method

In combinatorial mathematics, the stars and bars method is an ingenious way to find out how to distribute identical items into different groups. This technique is particularly useful when dealing with distribution problems with constraints.
Imagine you have a certain number of marks (or stars) to be given to questions (or groups). The stars represent the items we need to distribute, and the bars represent the dividers that separate items into different groups. For instance, distributing 14 marks among 8 questions involves arranging these marks (stars) along with the dividers (bars) that distinguish between the groups.

• You need 7 bars to create 8 groups, which are the questions here.
• Adding these 7 bars to the 14 stars gives us 21 total symbols to arrange.

This arrangement can be mathematically represented by the formula \( \binom{n + k - 1}{k - 1} \), where \( n \) is the number of stars and \( k \) is the number of groups. Through this method, you can neatly calculate how many different ways there are to distribute the items.

Binomial Coefficient

The binomial coefficient is a central mathematical concept that shows up in the stars and bars method. It provides a way to calculate combinations, or how many ways you can choose items from a set.
The symbol \( \binom{n}{k} \) represents the number of ways to choose \( k \) elements from a set of \( n \) elements, ignoring order. What makes it powerful is its ability to count arrangements and combinations efficiently.

• This is calculated as \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n! \) denotes the factorial of \( n \).
• In our problem, using \( \binom{21}{7} \), we determine the number of ways to arrange 14 stars and 7 bars among the 21 positions available.

Mastering the use of the binomial coefficient allows you to tackle a wide range of distribution and combinatorial problems.

Constraints in Distribution

Constraints in distribution problems add an additional layer of complexity. Here, we don't just care about distributing marks; every question has to receive at least 2 marks.
By including this constraint, the problem changes from simply dividing marks to fulfilling a minimum requirement for each group before proceeding with the remaining items.
To handle these constraints, follow a strategic approach:

• First, fulfill the constraint by giving each question the necessary 2 marks. This is done right away, reducing the problem size and complexity.
• Then, work with the leftover marks (or resources) freely, as the restrictions have been satisfied.

Such constraints are common in distribution problems and recognizing how to handle them efficiently is crucial for solving combinatorial challenges. Using these methods, you can ensure every condition in the problem is met, leading to a successful solution.