Sum of the coefficients of the terms of degree m in the expansion of `(1 + x)^(n)(1 + y)^(n) (1 + z)^(n)` is

Options

A. `(""^(n)C_(m))^(3)`
B. `3(""^(n)C_(m))`
C. `""^(n)C_(3m)`
D. `""^(3n)C_(m)`

Answer

To find the sum of the coefficients of the terms of degree \( m \) in the expansion of \( (1 + x)^{n}(1 + y)^{n}(1 + z)^{n} \), we can follow these steps:

### Step 1: Understand the Expansion
The expression \( (1 + x)^{n}(1 + y)^{n}(1 + z)^{n} \) can be expanded using the binomial theorem. Each binomial expansion contributes terms of the form \( \binom{n}{k} x^k \), \( \binom{n}{j} y^j \), and \( \binom{n}{l} z^l \) for \( k, j, l \) being the respective powers of \( x, y, z \).

### Step 2: Identify the Degree Condition
We need to find the sum of the coefficients of terms where the total degree \( k + j + l = m \). This means we are looking for all combinations of \( k, j, l \) such that their sum equals \( m \).

### Step 3: Count the Combinations
The total number of ways to choose \( k, j, l \) such that \( k + j + l = m \) can be visualized as distributing \( m \) identical balls (the total degree) into 3 distinct boxes (the variables \( x, y, z \)). This is a combinatorial problem that can be solved using the "stars and bars" theorem.

### Step 4: Apply the Stars and Bars Theorem
According to the stars and bars theorem, the number of ways to distribute \( m \) identical items into \( r \) distinct groups is given by:
\[ \binom{m + r - 1}{r - 1} \]
In our case, \( r = 3 \) (for \( x, y, z \)), so we have:
\[ \binom{m + 3 - 1}{3 - 1} = \binom{m + 2}{2} \]

### Step 5: Multiply by the Coefficient from Each Expansion
Each of the \( k, j, l \) can be chosen from \( n \) items, so the coefficient for each selection is \( \binom{n}{k} \), \( \binom{n}{j} \), and \( \binom{n}{l} \). Thus, the sum of the coefficients of the terms of degree \( m \) is:
\[ \sum_{k+j+l=m} \binom{n}{k} \binom{n}{j} \binom{n}{l} \]

### Step 6: Use the Multinomial Theorem
The sum of the coefficients of the terms of degree \( m \) in the expansion can be simplified to:
\[ \binom{3n}{m} \]
This is because we are essentially choosing \( m \) items from a total of \( 3n \) items (since each variable contributes \( n \) items).

### Conclusion
Thus, the sum of the coefficients of the terms of degree \( m \) in the expansion of \( (1 + x)^{n}(1 + y)^{n}(1 + z)^{n} \) is:
\[ \boxed{\binom{3n}{m}} \]