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Question

A code word of 4 letters consists of two distinct consonants from the English alphabets followed by two digits from 1 to 9, with repetition allowed in digits. If the number of code words so formed ending with an even digit is 432×k, then k is equal to


  1. 7

  2. 35

  3. 49

  4. 5


Solution

The correct option is B

35


Number of code words ending with even digit

=21×20×9×4  [Because there are 21 consonants and the last place can be occupied by any one of 2,4,6,8]

432×k=21×20×9×4

k=35

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