If the parabola `y=a x^2+b x+c ,w h e r ea , b , c in {1,2,3,4,5,6}` pass through (1, 8), then number of such parabola is `k` the `k-12`

Answer

To solve the problem of finding the number of parabolas of the form \( y = ax^2 + bx + c \) that pass through the point (1, 8), where \( a, b, c \) are integers from the set \{1, 2, 3, 4, 5, 6\}, we can follow these steps:

### Step 1: Set up the equation
Since the parabola passes through the point (1, 8), we can substitute \( x = 1 \) and \( y = 8 \) into the equation:
\[ 8 = a(1)^2 + b(1) + c \]
This simplifies to:
\[ 8 = a + b + c \]

### Step 2: Determine the range of values for \( a, b, c \)
We know that \( a, b, c \) can take values from the set \{1, 2, 3, 4, 5, 6\}. Therefore, we need to find combinations of \( a, b, c \) such that their sum equals 8.

### Step 3: Analyze possible values for \( a \)
We will analyze each possible value of \( a \) from 1 to 6 and find corresponding values of \( b \) and \( c \) that satisfy the equation \( b + c = 8 - a \).

1. **If \( a = 1 \)**:
\[ b + c = 8 - 1 = 7 \]
Possible pairs \((b, c)\): (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 combinations.

2. **If \( a = 2 \)**:
\[ b + c = 8 - 2 = 6 \]
Possible pairs \((b, c)\): (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) → 5 combinations.

3. **If \( a = 3 \)**:
\[ b + c = 8 - 3 = 5 \]
Possible pairs \((b, c)\): (1, 4), (2, 3), (3, 2), (4, 1) → 4 combinations.

4. **If \( a = 4 \)**:
\[ b + c = 8 - 4 = 4 \]
Possible pairs \((b, c)\): (1, 3), (2, 2), (3, 1) → 3 combinations.

5. **If \( a = 5 \)**:
\[ b + c = 8 - 5 = 3 \]
Possible pairs \((b, c)\): (1, 2), (2, 1) → 2 combinations.

6. **If \( a = 6 \)**:
\[ b + c = 8 - 6 = 2 \]
Possible pairs \((b, c)\): (1, 1) → 1 combination.

### Step 4: Calculate total combinations
Now we sum all the combinations found:
\[ 6 + 5 + 4 + 3 + 2 + 1 = 21 \]
Thus, the total number of parabolas \( k = 21 \).

### Step 5: Find \( k - 12 \)
Finally, we compute:
\[ k - 12 = 21 - 12 = 9 \]

### Final Answer:
The value of \( k - 12 \) is \( \boxed{9} \).