How many test cases are required to test a specification that has three conditions using a decision table testing which is not collapsed? 1 credit [K2]

Question

Options

• A2
• B4
• C6
• D8

• A
  4% (2)
• B
  7% (4)
• C
  2% (1)
• D
  88% (50)

Explanation

With three binary conditions in a non-collapsed decision table, every possible True/False combination must be represented, giving 2³ = 8 test cases - one column per unique combination of condition outcomes.

Why the distractors fail:

• A (2): Far too few; this ignores the combinatorial nature of decision tables entirely.
• B (4): Correct for two conditions (2² = 4), not three - a classic off-by-one-condition mistake.
• C (6): Comes from multiplying linearly (2 × 3), but conditions combine exponentially, not additively.

Memory tip: Use the formula 2ⁿ, where n is the number of conditions. Each condition doubles the column count because it can independently be T or F. Three conditions = 2 × 2 × 2 = 8. The phrase "collapsed = fewer, full = 2ⁿ" can help you remember that "not collapsed" always means the full power-of-two count.

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