Pin

My pin# is 4 digits long, it's an even number and all of the digits (0-9 are the options) are different. Even knowing all of this, what chance do you have of guessing it in a single guess?

1 2520 \frac{1}{2520} 1 10000 \frac{1}{10000} 1 5040 \frac{1}{5040} 1 2000 \frac{1}{2000} 1 5000 \frac{1}{5000}

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1 solution

Chung Kevin
Sep 30, 2015

Since the pin# is even, the last digit must be 0, 2, 4, 6, or 8 -- 5 options.
Since all of the digits are distinct, that leaves 9 options for the 3rd digit, then 8 for the 2nd digit, and then 7 for the 1st digit. 5 × 9 × 8 × 7 = 2520 5 \times 9 \times 8 \times 7 = 2520 . So, the chance of guessing her pin correctly in one guess is still only 1 2520 \frac{1}{2520} .

That said, if she hadn't said anything about it, it would have been 1 1 0 4 = 1 10000 \frac{1}{10^4} = \frac{1}{10000} so it's a much weaker pin now. So don't tell people things about your pin/passwords!

According to your calculation 0 can be on the first place but you told that it is four digit so it shouldn't be on first place so ams should be 1÷(2296)

Rohan Jasani - 5 years, 3 months ago

I solved this problem as below. but didn't get the right answer. can you please show me my error? Case 1 when the last digit is 0 then there are 9 options for 1st digit, 8 options for 2nd and 7 options for 3rd digit. So there are 9 8 7 = 504 ways

Case 2 The last digit is not equal to 0 4 options for last digit 8 options for 1st digit 8 options for 2nd digit 7 options for 3rd digit hence 8 8 7*4= 1792 ways

therefore totally 1792+504=2296 ways

Charuka Bandara - 5 years, 3 months ago

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A pin number can start with 0. It is not a "4-digit number", it just consists of 4 digits.

Chung Kevin - 5 years, 3 months ago

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Oh.. now i've got it. thanks.

Charuka Bandara - 5 years, 3 months ago

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