A person goes in for an examination in which there are four papers with a maximum of m marks from each paper. The number of ways in which one can get 2m marks is

Options

A. `.^(2m+3)C_(3)`
B. `((1)/(3))(m+1)(2m^(2)+4m+1)`
C. `((1)/(3))(m+1)(2m^(2)+4m+3)`
D. None of these

Answer

To solve the problem, we need to find the number of ways a person can score a total of 2m marks across four papers, each having a maximum score of m marks. Let's denote the scores in the four papers as x, y, z, and w. Thus, we need to solve the equation:

\[ x + y + z + w = 2m \]

with the constraints:

\[ 0 \leq x, y, z, w \leq m \]

### Step 1: Calculate the total number of non-negative integer solutions without constraints

First, we can find the total number of non-negative integer solutions to the equation \( x + y + z + w = 2m \) using the stars and bars method. The formula for the number of solutions to the equation \( x_1 + x_2 + ... + x_k = n \) in non-negative integers is given by:

\[ \binom{n + k - 1}{k - 1} \]

In our case, \( n = 2m \) and \( k = 4 \) (since we have four papers). Thus, the total number of solutions is:

\[ \binom{2m + 4 - 1}{4 - 1} = \binom{2m + 3}{3} \]

### Step 2: Subtract the cases where one or more scores exceed m

Next, we need to subtract the cases where one or more of the scores exceed m. Let's consider the case where \( x > m \). We can set \( x' = x - (m + 1) \) (where \( x' \) is a non-negative integer). The equation then becomes:

\[ x' + y + z + w = 2m - (m + 1) = m - 1 \]

Now we need to find the number of non-negative integer solutions to this new equation. Using the stars and bars method again, we have:

\[ \binom{(m - 1) + 4 - 1}{4 - 1} = \binom{m + 2}{3} \]

Since the same logic applies for \( y, z, \) and \( w \) exceeding m, we multiply this result by 4 (for each variable):

\[ 4 \cdot \binom{m + 2}{3} \]

### Step 3: Apply the Inclusion-Exclusion Principle

Now, we need to apply the Inclusion-Exclusion Principle to account for cases where two or more variables exceed m. However, since we are only interested in the first case where one variable exceeds m, we can directly calculate the total valid solutions:

The total number of valid solutions is:

\[ \text{Total Solutions} - \text{Excess Solutions} = \binom{2m + 3}{3} - 4 \cdot \binom{m + 2}{3} \]

### Final Result

Thus, the final number of ways in which one can get 2m marks is:

\[ \binom{2m + 3}{3} - 4 \cdot \binom{m + 2}{3} \]