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Note that although the codewords under the two methods are different, the word lengths are the same. We have lengths of 2 bits for A, and 3 bits for B, C, D and E, giving an average length of
:<math>\frac{2\,\text{bits}\cdot(15) + 3\,\text{bits} \cdot (7+6+6+5)}{39\, \text{symbols}} \approx 2.62\,\text{bits per symbol,}</math>
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