Question

# Probability outcome: toddler triplets whose names are red, green and blue each have a coat whose colour is given by their name, but they aren't clear on which coat belongs to which of them. when they go out, red chooses a coat at random among the three coats, green chooses a coat at random among the remaining two coats and blue is left to take the remaining coat. how many ways are there for exactly one of the toddler triplets to end up with their coat? help is the answer 3!=6 i dont understand how n! could be the way to go about this, if n! is for If you want to see how many ways you can order a certain number of things: n! (n factorial)

Answer

12

- There are three coats, and each toddler can end up with the correct coat or one of the other two coats. Let's consider the cases in which each of the toddlers end up with their own coat.
- Case 1: Only Red ends up with their own coat. Green and blue must each pick one of the two other coats. This can happen in 2 ways for each of them, for a total of 2 x 2 = 4 ways.
- Case 2: Only Green ends up with their own coat. Red and blue each have two choices of coat. This can happen in 2 x 2 = 4 ways.
- Case 3: Only Blue ends up with their own coat. Red and Green each have two choices of coat. This can happen in 2 x 2 = 4 ways.
- Adding up the possibilities from each case, there are a total of 4 + 4 + 4 = 12 ways for exactly one of the toddler triplets to end up with their coat.