Weekend combinatorics (III)

2 Dec, 2022 at 13:11 | Posted in Statistics & Econometrics | 3 Comments

Combinatorics in .NET - Part I - Permutations, Combinations & Variations –  try { } catch { } meAn easy one this week: At a small economics conference, a photographer wants to line up nine participants for a photo. Two of them — Robert and Milton — insist on standing next to each other. How many different arrangements (lineups) are possible?

3 Comments

  1. ChatGPT’s first answer (subsequent answers were different):
    .
    《If Robert and Milton insist on standing next to each other, there are two possible arrangements for them: they can stand either at the beginning or at the end of the lineup. For each of these two arrangements, there are 8! (8 factorial, which is the product of all positive integers less than or equal to 8) ways to arrange the remaining participants in the lineup, since there are 8 participants left to arrange. Therefore, the total number of possible arrangements is 2 * 8! = 2 * 40,320 = 80,640.》
    .
    Not trusting this answer, how surprised was I when my numerical methods program returned the same result?
    .
    https://subbot.org/misc/perms_3.rb.txt

    • Why can Robert and Milton be only at the beginning or the end of the line? Can’t they stand together anywhere in the line? It seems that if they stand together, they can be considered as one unit in a group of 9. If so, then the answer would be 9!
      What am I missing?
      – – John Lounsbury

      • John, you’re absolutely right. The question doesn’t imply that Robert and Milton only can be at the beginning or the end of the lineup (although rsm seems to think so). We have 9 people, but only 8 ‘groups’ since R and M can’t be separated. So we have 8! — and then (this is where students often go wrong) we have to multiply with 2! since R and M can be both to the left and right of each other. rsm got the correct answer, but for the wrong reason 🙂


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