Did you solve it? A head of pin

Earlier today, I posed this puzzle to you about PINs, the four-digit passcodes we use for phones and bank accounts.

In the comments, many people said the answer was obvious. These people fell for the trap.

Confusing PINs

You and a friend discuss how you choose four -digit PIN codes. You establish that neither of you would ever use the number 0.

"I like to choose four different random numbers", you say .

"I like to choose three different random digits", they reply, "so one of the digits is used twice."

Which strategy gives the largest pool of four-digit PINs?

Answer: Surprisingly, both strategies create exactly the same number of possible PINs.

Show me how it works: your strategy is to choose four non-zero random digits different. There are nine choices for the first digit, eight for the second digit, seven for the third, and six for the last digit. The number of possible PINs is therefore 9 x 8 x 7 x 6 = 3024.

Your friend's strategy is to choose three different non-zero random numbers, thus repeating the 'one of them. There are 9 choices for the digit that is repeated. And there are 6 positions for repeated digits:

xx--

x-x-

x--x

-xx-

-x-x

--xx

There are 8 choices for the other leftmost digit, and 7 choices remain for the last digit. So the total number is again 9 x 8 x 7 x 6 = 3024.

By mathematical coincidence, the number of possibilities when choosing three or four numbers is exactly the same. Good!

Hope you enjoyed this week's brain teaser. I'll be back in two weeks.

I put a puzzle here every other week on a Monday. I'm always on the lookout for great puzzles. If you would like to suggest one, write to me.

I am the author of several math and puzzle books, and also the co - author with Ben Lyttleton of the Football School series of children's books. The latest in the Football School series is The Greatest Ever Quiz Book, available now!

I lecture on math and puzzles ( online and in person). If your school is interested, please contact us.