Chapter 10 Counting Methods and Probability - 10.4 Find Probabilities of Disjoint and Overlapping Events - 10.4 Exercises - Problem Solving - Page 712: 46

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The number of possible codes for 6 houses is: $4096^6=4,722,366,482,869,645,213,696$ The number of ways that 6 houses can have 6 distinct codes is: $_{4096}P_6=\frac{4096!}{4090!}=4,705,096,570,216,277,114,880$ The probability that at least 2 of the 6 houses have the same code: $P=1-\frac{4,705,096,570,216,277,114,880}{4,722,366,482,869,645,213,696}=0.003657$