Concept & Reasoning

Each flip-flop stores one binary bit, so n flip-flops form 2ⁿ unique bit patterns. A 4-bit counter, for example, runs from 0000₂ (0) to 1111₂ (15), i.e. 0 to 2⁴−1. This generalizes directly to n bits.

How to Solve It in the Exam

n bits → 2ⁿ states → maximum numeric count 2ⁿ−1.

Important Exam Point

Separate 'number of states' from 'largest count'.

Quick Trick

n bits → 2ⁿ states → maximum numeric count 2ⁿ−1.

Common Mistake

Using 2ⁿ as the maximum count instead of the number of states.

Concept & Reasoning

Each flip-flop stores one binary bit, so n flip-flops form 2ⁿ unique bit patterns. A 4-bit counter, for example, runs from 0000₂ (0) to 1111₂ (15), i.e. 0 to 2⁴−1. This generalizes directly to n bits.

How to Solve It in the Exam

n bits → 2ⁿ states → maximum numeric count 2ⁿ−1.

Important Exam Point

Separate 'number of states' from 'largest count'.

Quick Trick

n bits → 2ⁿ states → maximum numeric count 2ⁿ−1.

Common Mistake

Using 2ⁿ as the maximum count instead of the number of states.

Concept & Reasoning

Each flip-flop stores one binary bit, so n flip-flops form 2ⁿ unique bit patterns. A 4-bit counter, for example, runs from 0000₂ (0) to 1111₂ (15), i.e. 0 to 2⁴−1. This generalizes directly to n bits.

How to Solve It in the Exam

n bits → 2ⁿ states → maximum numeric count 2ⁿ−1.

Important Exam Point

Separate 'number of states' from 'largest count'.

Quick Trick

n bits → 2ⁿ states → maximum numeric count 2ⁿ−1.

Common Mistake

Using 2ⁿ as the maximum count instead of the number of states.