QQuestionMathematics
QuestionMathematics
The Khmer alphabet has 74 letters which includes 33 consonants, 23 vowels, and various diacritics. Further, the Khmer language has its own set of numerals representing $0,1,2,3,4,5,6,7,1$ and 9 . If any five letters of the Khmer alphabet followed three numerals by could be used to form an eightcharacter code, how many codes would have at least one repeated letter or one repeated numeral?
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Answer
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Step 1:Let's solve this step by step:
Step 2:: Understand the components
- Total letters in Khmer alphabet: 74 - Total numerals: 10 (0 - 9) - Code structure: 5 letters followed by 3 numerals - Total code length: 8 characters
Step 3:: Calculate total possible codes
- Total possible codes: $$\binom{74}{5} \times \binom{10}{3}
- Numeral selection: \binom{10}{3}
Step 4:: Calculate codes with no repeats
- Total codes without repeats: $$\binom{74}{5} \times \binom{10}{3}
- Numeral selections without repeats: \binom{10}{3}
Step 5:: Use complementary counting
- $$\binom{74}{5} \times \binom{10}{3} - \binom{74}{5} \times \binom{10}{3}
- Total codes with at least one repeat = Total possible codes - Codes without repeats
Final Answer
The number of codes with at least one repeated letter or numeral is \binom{74}{5} \times \binom{10}{3}.
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