Ask Question
7 April, 13:29

Question 1 (Module Outcome #1) : A company wishes to offer employees a choice for their personal identification numbers (PIN). Each pin must be a strong of length 4. The first and second characters can be any digit 0, ..., 9; the third and fourth character can be any of the 5 symbols chosen from {@,#,!,%,&}. No repeated characters are permitted. How many PINs are possible?

+4
Answers (1)
  1. 7 April, 13:54
    0
    1800

    Step-by-step explanation:

    Since the first and second characters can be any digit 0, ..., 9

    So, there are 10 possible digits for first place

    thus we have only 9 digits possible for second place

    Also, it is given that the third and fourth character must be any of 5 given symbols

    So, we have 5 possible symbols for third place

    and since no repetition is allowed we have only 4 symbols left for the last place.

    ∴ Number of possible PINs are:

    10 * 9 * 5 * 4 = 1800
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Question 1 (Module Outcome #1) : A company wishes to offer employees a choice for their personal identification numbers (PIN). Each pin ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers