Find the cofficient of `x^2 y^5` in the expansion of `(3+2x-y)^(10)`

Answer

To find the coefficient of \( x^2 y^5 \) in the expansion of \( (3 + 2x - y)^{10} \), we will use the multinomial expansion formula. The general term in the expansion of \( (a + b + c)^n \) is given by:

\[ T = \frac{n!}{p!q!r!} a^p b^q c^r \]

where \( n = p + q + r \).

### Step 1: Identify the terms
In our case, we have:
- \( a = 3 \)
- \( b = 2x \)
- \( c = -y \)
- \( n = 10 \)

We want to find the coefficient of \( x^2 y^5 \). This means we need to identify \( p \), \( q \), and \( r \) such that:
- \( b^q = (2x)^q \) gives \( x^2 \) (thus \( q = 2 \))
- \( c^r = (-y)^r \) gives \( y^5 \) (thus \( r = 5 \))
- The remaining term \( a^p = 3^p \) will be determined by \( p = n - q - r \)

### Step 2: Calculate \( p \)
Now, we calculate \( p \):
\[ p = n - q - r = 10 - 2 - 5 = 3 \]

### Step 3: Calculate the coefficient
Now we can substitute \( p \), \( q \), and \( r \) into the multinomial coefficient formula:

\[ \text{Coefficient} = \frac{10!}{p!q!r!} \cdot a^p \cdot b^q \cdot c^r \]

Substituting the values:
\[ \text{Coefficient} = \frac{10!}{3!2!5!} \cdot 3^3 \cdot (2x)^2 \cdot (-y)^5 \]

Calculating each part:
- \( 10! = 3628800 \)
- \( 3! = 6 \)
- \( 2! = 2 \)
- \( 5! = 120 \)

Calculating the multinomial coefficient:
\[ \frac{10!}{3!2!5!} = \frac{3628800}{6 \cdot 2 \cdot 120} = \frac{3628800}{1440} = 2520 \]

Now calculate \( 3^3 \):
\[ 3^3 = 27 \]

Calculate \( (2x)^2 \):
\[ (2x)^2 = 4x^2 \]

Calculate \( (-y)^5 \):
\[ (-y)^5 = -y^5 \]

### Step 4: Combine everything
Now we combine everything:
\[ \text{Coefficient} = 2520 \cdot 27 \cdot 4 \cdot (-1) \]

Calculating:
\[ 2520 \cdot 27 = 68040 \]
\[ 68040 \cdot 4 = 272160 \]
Since we have a negative sign from \( (-y)^5 \):
\[ \text{Coefficient} = -272160 \]

### Final Answer
Thus, the coefficient of \( x^2 y^5 \) in the expansion of \( (3 + 2x - y)^{10} \) is:
\[ \text{Coefficient} = -272160 \]