Question

The number of cubic equations that can be formed with the coefficients 0, 3, 2, 4, 5, 6 when repetition of coefficients is allowed, is....

• A. \(720\)
• B. \(1296\)
• C. \(864\)
• D. \(1080\)

Hint:

The leading coefficient cannot be zero, so fill that place first and the remaining places freely.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A cubic equation has the form \(ax^3 + bx^2 + cx + d = 0\) with four coefficients. For it to be cubic, \(a \neq 0\). The other three can be any of the six numbers because repetition is allowed.

Step 2: Key Formula or Approach:
Use the fundamental counting principle: multiply the number of choices for each coefficient.

Step 3: Detailed Explanation:
Choices for \(a\): any of the 5 non-zero numbers (3, 2, 4, 5, 6).
Choices for each of \(b\), \(c\), \(d\): 6 each.
\[ \text{Total} = 5 \times 6 \times 6 \times 6 = 5 \times 216 = 1080 \]

Step 4: Why the other options are wrong.
\(6^4 = 1296\) counts the cases with \(a = 0\), which are not cubic. 720 and 864 do not come from any valid counting here.

Final Answer:
The number of cubic equations is 1080, option (D). \[ \boxed{1080} \]