So I was trying to think about a problem to upload to Brilliant, and I came across a kind of difficult one that I'm not sure what's the right answer for. Try to arrange all the digits from 0-9(that's ten total) so that:
Every digit cannot be placed at the same place as its number (assuming we count places starting from 0), for example 0 can't be placed at the place 0.
As it concludes from the problem, no repetition is allowed. If I wrote 3 I can't choose it again.
Allowed arrangement: 9012345678, 1234567890
Digits representing places: 0123456789 0123456789
Explanation: As you can see, no digit in the arrangement is similar to the digit representing it place and no digit is repeated.
Forbidden arrangement: 0987654321,1123456789
Digits representing places: 0123456789 , 0 123456789
Explanation for the first number: 0 is placed at the place 0. Explanation for the second number: 1 appears more than once., 1 is located in the place 1.
The question: How many "allowed" arraignments can we make?
So my problem is that I have written a computer program using C# and I got the number 1,320,128, but the calculation I made by solving it (using combinatorics principles) brought me to 1,632,960. I prefer not to write here how exactly I got this numbers to encourage further examination and to prevent us from choosing similar ways and having similar mistakes. So I really hope at least one of my results is correct, and I can't wait to see how would you solve it! By the way, after we will confirm the solution and result I will post this problem and solution with credit to the solvers. Good luck and thank you very much!