# Intuition behind permutations + probability

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?

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?

• 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 – lkdlst Apr 22 at 10:16
• $\frac{1}{10!}$ 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 – Henry Apr 22 at 10:24
• 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? – lkdlst Apr 22 at 10:28
• You might want to read about rencontres numbers – Henry Apr 22 at 10:57