Question

How many ways can an examiner assign 90 points to 12 questions with each question getting at least four points?

Short answer

There are 178,462,987 ways to assign 90 points to 12 questions.

Step-by-step solution

Understanding the Problem

The examiner wants to allocate 90 points to 12 questions, with each question getting at least 4 points. This means initially, we allocate 4 points to each question.

Initial Allocation of Points

Since each question receives at least 4 points, for 12 questions, we'll allocate a total of \(4 \times 12 = 48\) points at first. This leaves us with \(90 - 48 = 42\) points to distribute freely among the 12 questions.

Formulate the Remaining Problem

The problem now is to determine the number of ways to distribute 42 points among 12 questions. This is a typical 'stars and bars' combinatorial problem where we need to find the non-negative integer solutions to the equation \(x_1 + x_2 + \cdots + x_{12} = 42\).

Apply Stars and Bars Theorem

In combinatorics, the number of solutions to \(x_1 + x_2 + \ldots + x_{n} = k\) is given by the formula \(\binom{n+k-1}{n-1}\). Here, \(n = 12\) and \(k = 42\). Thus, the number of ways to distribute the points is \(\binom{42+12-1}{12-1} = \binom{53}{11}\).

Compute the Combination

Calculate the combination \(\binom{53}{11}\). This is computed as \(\frac{53!}{11! \cdot 42!}\), which represents the number of ways to choose 11 positions for the bars out of 53 total slots (42 stars plus 11 bars).

Final Calculation

After calculating \(\frac{53!}{11! \cdot 42!}\), we find that the number of ways to assign the extra 42 points is 178,462,987.

Key concepts

Stars and Bars Theorem

The Stars and Bars Theorem is a fundamental concept in combinatorics used to solve problems involving the distribution of a certain number of identical items (stars) into distinct groups (bins).
The approach typically involves transforming the problem into finding the non-negative integer solutions to an equation.
In our context, envision stars as the additional points that need to be distributed, and the bars as the separators between different questions.

To visualize, imagine placing these stars in a line to represent points and using bars to divide them into sections for each question.

• If we have 42 extra points (stars) to distribute into 12 questions (bins), we need 11 bars to create the separation between these sections.
• The configuration of stars and bars can be done in various combinations, and the number of possible combinations gives us the number of ways points can be distributed.

The formula derived from the Stars and Bars Theorem helps us find the total number of ways by arranging stars and bars in a line, completely solving the distribution issue without having to list out each possible way manually.

Integer Solutions

Finding integer solutions is pivotal in combinatorial problems involving distribution, like in our exam points scenario.
When tasked with distributing items, we convert our problem to finding how many ways we can assign whole number values to variables that sum up to a total.

In simpler terms, each variable represents a group (or a question in our scenario) to which points are allocated. For example, in distributing 42 points across 12 questions, we define variables such as \( x_1, x_2, \ldots, x_{12} \) where each is a non-negative integer indicating points assigned to that specific question.
The equation \( x_1 + x_2 + \ldots + x_{12} = 42 \) ensures every possible allocation of points is covered by all potential combinations of integer solutions.

• This framework allows us to systematically tackle more complex distribution problems without overlooking any scenario.
• It underscores the underlying mathematical fabric enabling the exact counting of combinations or arrangements, much like efficiently organizing elements under specific constraints.

Point Distribution

Point distribution problems often appear in contexts where allocation under certain conditions is required.
In the exercise example, we had a total of 90 points that needed initial allocation followed by free distribution among questions, respecting the condition that each received at least four points.

The problem initially assigns fixed points to ensure each question meets a minimum requirement, using a direct calculation: allocating \(4 \times 12 = 48\) points.

• This simplification trims down the problem to distributing the remaining 42 points more flexibly among the 12 questions.
• Such a process systematically reduces complexity, focusing only on the unrestricted portion of the distribution task.

To solve, we use the stars (points) and bars (dividers) to exhaustively consider every possible way.
This structured approach ensures no allocation possibility is overlooked, leveraging mathematical tools to simplify what might otherwise become a cumbersome counting exercise.