0

It is a follow up to my previous question here The only difference is 12 digits numbers are picked from 14 numbers.

Thanks to many great people, I have understood how to solve the previous question.

However, if I am to pick 12 numbers out of the total 14 numbers, many new cases emerge when I having 0, or 2, or 4 in the end there are a different number of them left to be choose, I cannot just simply remove the repetitive case using division. What can I do?

Kevin
  • 365
  • 1
  • 10
  • If you truly want to generalize things... I would probably go about this with generating functions or with inclusion-exclusion. You are correct that simply dividing is not allowable here as we overcounted certain arrangements a potentially different number of times each. – JMoravitz Sep 11 '19 at 14:17

1 Answers1

1

There are $20$ different ways to cross out two digits, namely $$01,02,03,04,05,11,12,13,14,15,22,23,24,25,33,34,35,44,45,55\ .$$ For each of these solve the old problem, and add up. (The $20$ cases reduce a little by remarking that the problem is symmetric in $3$ and $5$.)

Christian Blatter
  • 226,825