The number of ways in which an examiner can assign 40 marks to 6 questions, giving not less than 3 marks to any question, is

Options

A. `""^(27)C_(22)`
B. `""^(27)C_(21)`
C. `""^(26)C_(22)`
D. none of these

Answer

To solve the problem of assigning 40 marks to 6 questions with the condition that each question receives at least 3 marks, we can follow these steps:

### Step 1: Adjust the Marks
Since each of the 6 questions must receive at least 3 marks, we first allocate 3 marks to each question. This means we will initially distribute:
\[ 3 \times 6 = 18 \text{ marks} \]
Thus, we have:
\[ 40 - 18 = 22 \text{ marks left to distribute} \]

### Step 2: Define the New Variables
Let \( m_i \) be the marks assigned to question \( i \) (where \( i = 1, 2, 3, 4, 5, 6 \)). After assigning the minimum marks, we can redefine the marks for each question as:
\[ m_i' = m_i - 3 \]
This means \( m_i' \) can be zero or more (i.e., \( m_i' \geq 0 \)). The equation now becomes:
\[ m_1' + m_2' + m_3' + m_4' + m_5' + m_6' = 22 \]

### Step 3: Use the Stars and Bars Theorem
Now, we need to find the number of non-negative integer solutions to the equation:
\[ m_1' + m_2' + m_3' + m_4' + m_5' + m_6' = 22 \]
According to the stars and bars theorem, the number of ways to distribute \( n \) indistinguishable objects (marks) into \( r \) distinguishable boxes (questions) is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 22 \) and \( r = 6 \):
\[ \text{Number of ways} = \binom{22 + 6 - 1}{6 - 1} = \binom{27}{5} \]

### Step 4: Calculate the Binomial Coefficient
Now we compute:
\[ \binom{27}{5} = \frac{27!}{5!(27 - 5)!} = \frac{27!}{5! \cdot 22!} \]
Calculating this gives:
\[ \binom{27}{5} = \frac{27 \times 26 \times 25 \times 24 \times 23}{5 \times 4 \times 3 \times 2 \times 1} = 142506 \]

### Final Answer
Thus, the number of ways in which an examiner can assign 40 marks to 6 questions, giving not less than 3 marks to any question, is:
\[ \boxed{142506} \]