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?

Question

Mathematics

Angie Green

23 October, 21:58

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?

+3

Answers (1)

Know the Answer?

Topaz · Answer 1

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