Find the coefficient of `x^3y^4z^5` in the expansion of `(xy+yz+zx)^6`

Options

A. 120
B. 20
C. 30
D. 60

Explanation

To find the coefficient of \( x^3 y^4 z^5 \) in the expansion of \( (xy + yz + zx)^6 \), we can follow these steps:

### Step 1: Identify the terms in the expansion
The expression \( (xy + yz + zx)^6 \) can be expanded using the multinomial theorem. The general term in the expansion can be expressed as:
\[ \frac{6!}{r! s! t!} (xy)^r (yz)^s (zx)^t \]
where \( r + s + t = 6 \).

### Step 2: Determine the powers of \( x, y, z \)
We want the term that contains \( x^3 y^4 z^5 \). From the general term, we can express the powers of \( x, y, z \) in terms of \( r, s, t \):
- The power of \( x \) is \( r + t \)
- The power of \( y \) is \( r + s \)
- The power of \( z \) is \( s + t \)

We need to set up the following equations based on the desired powers:
1. \( r + t = 3 \) (for \( x^3 \))
2. \( r + s = 4 \) (for \( y^4 \))
3. \( s + t = 5 \) (for \( z^5 \))

### Step 3: Solve the system of equations
We have the following equations:
1. \( r + t = 3 \) (1)
2. \( r + s = 4 \) (2)
3. \( s + t = 5 \) (3)

From equation (1), we can express \( t \) as:
\[ t = 3 - r \]

Substituting \( t \) into equation (3):
\[ s + (3 - r) = 5 \implies s - r = 2 \implies s = r + 2 \]

Now substitute \( s \) into equation (2):
\[ r + (r + 2) = 4 \implies 2r + 2 = 4 \implies 2r = 2 \implies r = 1 \]

Now substituting \( r = 1 \) back to find \( s \) and \( t \):
\[ s = r + 2 = 1 + 2 = 3 \]
\[ t = 3 - r = 3 - 1 = 2 \]

Thus, we have:
- \( r = 1 \)
- \( s = 3 \)
- \( t = 2 \)

### Step 4: Calculate the coefficient
Now we can find the coefficient using the multinomial coefficient:
\[ \text{Coefficient} = \frac{6!}{r! s! t!} = \frac{6!}{1! 3! 2!} \]

Calculating this:
\[ 6! = 720, \quad 1! = 1, \quad 3! = 6, \quad 2! = 2 \]
\[ \text{Coefficient} = \frac{720}{1 \times 6 \times 2} = \frac{720}{12} = 60 \]

### Final Answer
The coefficient of \( x^3 y^4 z^5 \) in the expansion of \( (xy + yz + zx)^6 \) is \( 60 \).

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