The number of ways in which an examiner can assign 50 marks to 10 questions giving not less than 3 marks to any question is :

Options

A. `"^(29)C_9`
B. `"^(47)C_3`
C. `"^(52)C_2`
D. `"^(40)C_(10)`

Answer

To solve the problem of assigning 50 marks to 10 questions with the condition that each question receives at least 3 marks, we can follow these steps:

### Step 1: Understand the Problem
We need to assign marks to 10 questions such that the total marks assigned is 50, and each question receives at least 3 marks.

### Step 2: Assign Minimum Marks
Since each of the 10 questions must receive at least 3 marks, we can start by assigning 3 marks to each question.

- Marks assigned to each question = 3
- Total marks assigned = 10 questions × 3 marks = 30 marks

### Step 3: Calculate Remaining Marks
After assigning the minimum marks, we calculate how many marks are left to be distributed.

- Total marks to be distributed = 50 marks
- Marks already assigned = 30 marks
- Remaining marks = 50 - 30 = 20 marks

### Step 4: Set Up the New Equation
Now, we need to distribute the remaining 20 marks among the 10 questions. We can denote the additional marks given to each question as \( y_1, y_2, y_3, \ldots, y_{10} \).

The new equation becomes:
\[ y_1 + y_2 + y_3 + \ldots + y_{10} = 20 \]

### Step 5: Apply the Stars and Bars Theorem
To find the number of non-negative integer solutions to the equation \( y_1 + y_2 + y_3 + \ldots + y_{10} = 20 \), we can use the "Stars and Bars" theorem. According to this theorem, the number of ways to distribute \( n \) indistinguishable objects (stars) into \( r \) distinguishable boxes (questions) is given by the formula:
\[ \binom{n + r - 1}{r - 1} \]

In our case:
- \( n = 20 \) (the remaining marks)
- \( r = 10 \) (the questions)

### Step 6: Substitute Values into the Formula
Now we substitute \( n \) and \( r \) into the formula:
\[ \text{Number of ways} = \binom{20 + 10 - 1}{10 - 1} = \binom{29}{9} \]

### Step 7: Conclusion
Thus, the number of ways in which the examiner can assign 50 marks to 10 questions, giving not less than 3 marks to any question, is \( \binom{29}{9} \).