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EXERCISE 3.3 · Q78

Q.A code word is formed by two different English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.

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Two different letters from the 26-letter alphabet, ordered, can be chosen in 26P2=26×25=650{}^{26}P_2 = 26\times25=650 ways. Two distinct non-zero digits (from 1–9, 9 digits), ordered, can be chosen in 9P2=9×8=72{}^9P_2=9\times8=72 ways. By the Multiplication Principle, the total number of code words is 650×72=46800650\times72=46800. For code words ending in an even digit: the four even non-zero digits are 2,4,6,8, so the last digit has 4 choices; the second-last digit (still a non-zero digit, distinct from the last) has 8 re …

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