Find the coefficient of `a^(5) b^(5)c^(5)d^(6)` in the expansion of the following expression. `(bcd + acd + abd+ abc)^(7)`

Answer

To find the coefficient of \( a^5 b^5 c^5 d^6 \) in the expansion of \( (bcd + acd + abd + abc)^7 \), we can follow these steps:

### Step 1: Identify the terms in the expression
The expression is \( (bcd + acd + abd + abc)^7 \). We can rewrite this as:
\[ (bcd + acd + abd + abc)^7 \]

### Step 2: Use the multinomial theorem
According to the multinomial theorem, the expansion of \( (x_1 + x_2 + x_3 + x_4)^n \) can be expressed as:
\[ \sum \frac{n!}{k_1! k_2! k_3! k_4!} x_1^{k_1} x_2^{k_2} x_3^{k_3} x_4^{k_4} \]
where \( k_1 + k_2 + k_3 + k_4 = n \).

### Step 3: Set up the equation for the powers
We want to find the coefficient of \( a^5 b^5 c^5 d^6 \). This means we need to select terms such that the total powers of \( a, b, c, \) and \( d \) match \( 5, 5, 5, \) and \( 6 \) respectively.

### Step 4: Determine the combinations
We can express the terms in the expansion as:
- \( bcd \) contributes \( 0 \) to \( a \), \( 1 \) to \( b \), \( 1 \) to \( c \), and \( 1 \) to \( d \)
- \( acd \) contributes \( 1 \) to \( a \), \( 0 \) to \( b \), \( 1 \) to \( c \), and \( 1 \) to \( d \)
- \( abd \) contributes \( 1 \) to \( a \), \( 1 \) to \( b \), \( 0 \) to \( c \), and \( 1 \) to \( d \)
- \( abc \) contributes \( 1 \) to \( a \), \( 1 \) to \( b \), \( 1 \) to \( c \), and \( 0 \) to \( d \)

Let:
- \( x_1 \): number of times \( bcd \) is chosen
- \( x_2 \): number of times \( acd \) is chosen
- \( x_3 \): number of times \( abd \) is chosen
- \( x_4 \): number of times \( abc \) is chosen

We need to solve the equations:
1. \( x_1 + x_2 + x_3 + x_4 = 7 \) (total selections)
2. \( x_2 + x_3 + x_4 = 5 \) (for \( b \))
3. \( x_1 + x_2 + x_4 = 5 \) (for \( c \))
4. \( x_1 + x_2 + x_3 = 6 \) (for \( d \))

### Step 5: Solve the equations
From the equations, we can derive:
- From equation 2: \( x_1 + 7 - (x_2 + x_3 + x_4) = 2 \) implies \( x_1 = 2 \)
- From equation 3: \( x_2 + x_4 = 3 \)
- From equation 4: \( x_3 = 1 \)

Now substituting \( x_1 = 2 \) and \( x_3 = 1 \) into the total selections gives:
\[ 2 + x_2 + 1 + x_4 = 7 \implies x_2 + x_4 = 4 \]
Now we have:
- \( x_2 + x_4 = 3 \)
- \( x_2 + x_4 = 4 \)

This leads to \( x_2 = 3 \) and \( x_4 = 1 \).

### Step 6: Calculate the coefficient
Now we have:
- \( x_1 = 2 \)
- \( x_2 = 3 \)
- \( x_3 = 1 \)
- \( x_4 = 1 \)

The coefficient is given by:
\[ \frac{7!}{x_1! x_2! x_3! x_4!} = \frac{7!}{2! 3! 1! 1!} = \frac{5040}{2 \times 6 \times 1 \times 1} = \frac{5040}{12} = 420 \]

### Final Answer
The coefficient of \( a^5 b^5 c^5 d^6 \) in the expansion of \( (bcd + acd + abd + abc)^7 \) is \( 420 \).