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I'm new to stats so I'm struggling to grasp the intuition behind permutations + probability. Would anybody be able to help me with parts (b) and (c) of this question?

Any help is greatly appreciated :)

A game requires you to match 10 words to 10 images (where each word correctly labels only one image). A machine randomly matches each word to a different image, where all possible labellings are equally likely.

What is the probability that the machine matches:

  • (a) all 10 words correctly?
  • (b) the first 7 words correctly?
  • (c) exactly 9 of the 10 words correctly?

This question is from a stats class problem set.

My work so far:

  • (a) $\frac{1}{10!}$ as $10!$ permutations are possible with only 1 having all correctly matched.
  • (b) I think $\frac{10!}{3!}$ but not 100% sure.
  • (c) Not too sure where to begin but I realise that if the last word is incorrectly matched then one of the 9 must also be incorrectly matched. What is the question instead said exactly 8 correct?
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Hints:

  • How many equally likely permutations are there?

  • How many permutations get all $10$ words correct? What proportion of the total is this?

  • How many permutations get the first $7$ words correct and the last $3$ in any order? What proportion of the total is this?

  • If exactly $9$ out of $10$ words are correct then $1$ word is matched to a wrong image. What might have happened to the other word that should have been matched to that image?

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  • $\begingroup$ Thanks! Here's what I've got so far. • $10!$ permutations possible • $\frac{1}{10!}$ permutations gets all 10 correct • $\frac{10!}{7!}$ get the first 7 correct? • Not too sure for the last one $\endgroup$ – lkdlst Apr 22 at 10:16
  • $\begingroup$ $\frac{1}{10!}$ is correct. $\frac{10!}{7!}$ is not: if the first seven are correct, then you are counting the number of permutations of the last three. The final question has two possible answers: either the other word was matched correctly or it was matched incorrectly, and each of those has logical implications $\endgroup$ – Henry Apr 22 at 10:24
  • $\begingroup$ Right so $\frac{10!}{7!}$ should be $\frac{10!}{3!}$ and for the last one if it was matched incorrectly then that means it isn't possible to have the other 9 be correct - what if the question was instead asking about matching exactly 8 out of 10 correctly? $\endgroup$ – lkdlst Apr 22 at 10:28
  • $\begingroup$ You might want to read about rencontres numbers $\endgroup$ – Henry Apr 22 at 10:57

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