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 use the multinomial expansion. Here’s a step-by-step solution:

### Step 1: Identify the terms in the expansion
The expression can be rewritten as:
\[ (1 + x + (-x^2))^5 \]
We need to find the coefficient of \( x^4 \) in this expansion.

### Step 2: Use the multinomial theorem
According to the multinomial theorem, the expansion of \( (a + b + c)^n \) can be expressed as:
\[ \sum \frac{n!}{k_1! k_2! k_3!} a^{k_1} b^{k_2} c^{k_3} \]
where \( k_1 + k_2 + k_3 = n \). In our case, \( a = 1 \), \( b = x \), and \( c = -x^2 \), with \( n = 5 \).

### Step 3: Determine combinations that yield \( x^4 \)
We need to find all combinations of \( k_1 \), \( k_2 \), and \( k_3 \) such that:
\[ k_2 + 2k_3 = 4 \]
and
\[ k_1 + k_2 + k_3 = 5 \]

### Step 4: Solve the equations
From \( k_1 + k_2 + k_3 = 5 \), we can express \( k_1 \) as:
\[ k_1 = 5 - k_2 - k_3 \]
Substituting into the first equation:
\[ k_2 + 2k_3 = 4 \]
We can analyze possible values for \( k_3 \):

1. **If \( k_3 = 0 \)**:
- \( k_2 = 4 \)
- \( k_1 = 1 \)
- Contribution: \( \frac{5!}{1!4!0!} (1)^1 (x)^4 (-x^2)^0 = 5 \)

2. **If \( k_3 = 1 \)**:
- \( k_2 + 2(1) = 4 \) ⇒ \( k_2 = 2 \)
- \( k_1 = 5 - 2 - 1 = 2 \)
- Contribution: \( \frac{5!}{2!2!1!} (1)^2 (x)^2 (-x^2)^1 = -30 \)

3. **If \( k_3 = 2 \)**:
- \( k_2 + 2(2) = 4 \) ⇒ \( k_2 = 0 \)
- \( k_1 = 5 - 0 - 2 = 3 \)
- Contribution: \( \frac{5!}{3!0!2!} (1)^3 (x)^0 (-x^2)^2 = 15 \)

### Step 5: Sum the contributions
Now, we sum the contributions:
\[ 5 + (-30) + 15 = -10 \]

### Step 6: Find the absolute value
The coefficient \( k \) is \( -10 \), thus:
\[ |k| = 10 \]

### Final Answer
The absolute value of the coefficient \( |k| \) is:
\[ \boxed{10} \]