Coefficient of `x^4` in the expansion `(1 + x - x^2)^5` is k then `|k|` .

Explanation

To find the coefficient of \( x^4 \) in the expansion of \( (1 + x - x^2)^5 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (1 + x - x^2)^5 \]
We can group the terms as:
\[ (1 + x + (-x^2))^5 \]

### Step 2: Use the multinomial expansion
The multinomial expansion states that:
\[ (a_1 + a_2 + a_3)^n = \sum \frac{n!}{k_1! k_2! k_3!} a_1^{k_1} a_2^{k_2} a_3^{k_3} \]
where \( k_1 + k_2 + k_3 = n \). Here, we have:
- \( a_1 = 1 \)
- \( a_2 = x \)
- \( a_3 = -x^2 \)
- \( n = 5 \)

### Step 3: Find the relevant terms
We need to find the combinations of \( k_1, k_2, k_3 \) such that:
\[ k_2 + 2k_3 = 4 \]
and
\[ k_1 + k_2 + k_3 = 5 \]

### Step 4: Solve for \( k_1, k_2, k_3 \)
From the second equation, we can express \( k_1 \) as:
\[ k_1 = 5 - k_2 - k_3 \]
Substituting this into the first equation gives:
\[ k_2 + 2k_3 = 4 \]
Now we can try different values for \( k_3 \):
1. **If \( k_3 = 0 \)**:
\[ k_2 = 4 \quad \Rightarrow \quad k_1 = 1 \]
Contribution:
\[ \frac{5!}{1!4!0!} \cdot 1^1 \cdot x^4 \cdot (-x^2)^0 = 5 \]

2. **If \( k_3 = 1 \)**:
\[ k_2 + 2(1) = 4 \quad \Rightarrow \quad k_2 = 2 \quad \Rightarrow \quad k_1 = 2 \]
Contribution:
\[ \frac{5!}{2!2!1!} \cdot 1^2 \cdot x^2 \cdot (-x^2)^1 = \frac{120}{4} \cdot (-1) = -30 \]

3. **If \( k_3 = 2 \)**:
\[ k_2 + 2(2) = 4 \quad \Rightarrow \quad k_2 = 0 \quad \Rightarrow \quad k_1 = 1 \]
Contribution:
\[ \frac{5!}{1!0!2!} \cdot 1^1 \cdot x^0 \cdot (-x^2)^2 = \frac{120}{2} = 60 \]

### Step 5: Sum the contributions
Now, we sum all contributions:
\[ 5 - 30 + 60 = 35 \]

### Step 6: Find the absolute value
Thus, the coefficient \( k \) is \( 35 \), and we need \( |k| \):
\[ |k| = 35 \]

### Final Answer
\[ \boxed{35} \]