4

Jack has seven unlabelled seeds for different herbs (coriander, basil, parsley, sage, thyme, oregano and mint). He plants the seeds in one row. What is the probability that the coriander and basil are on the same side of the parsley?

I have looked at this and I'm wondering if the probabilities are the same based on the reasoning given?

binky
  • 37
  • Yes, the reasoning is exactly the same. – Parcly Taxel Oct 12 '20 at 09:10
  • The reasoning is the same, but this time, you have seven unlabelled seeds instead of six. Can you apply the reasoning shown there to your question? – Toby Mak Oct 12 '20 at 09:10
  • @TobyMak I'm struggling to understand how the questions differ. There are still only 3! ways of ordering the coriander, basil and parsley but it doesn't seem right that there should only be 6 different perumutations. – binky Oct 12 '20 at 09:16
  • @binky In that case, I was wrong: the questions don't differ at all. There are still $6$ orderings for each combination of sage, thyme, oregano and mint. – Toby Mak Oct 12 '20 at 09:21
  • @TobyMak Although the answers might be the same, I don't understand the logic behind fixing the four sage, thyme, oregano and mint herbs and only permutating the coriander, basil and parsley. Could you please elaborate on how they result in the same probability although there are now 7 items and not 6 (linked question). – binky Oct 12 '20 at 09:25
  • @ParclyTaxel Could you maybe help elaborate also? Thanks. – binky Oct 12 '20 at 09:55

1 Answers1

2

The reasoning in the linked question (or rather, the main answer in the linked question) applies equally to this problem.

All permutations of seeds can be grouped into sets of $3!=6$ that differ only in the positions of the coriander, basil and parcly parsley. In exactly four of those six permutations, for every set, the desired property is satisfied. Thus the probability must be $\frac23$.

Parcly Taxel
  • 103,344