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 1

There are 3! (3 factorial) ways for the toddler triplets to each choose a coat. However, we are only interested in the ways where exactly one of them ends up with their coat.

There are 3 ways for the first toddler to get their coat, 2 ways for the second toddler to get their coat, and 1 way for the third toddler to get their coat.

Therefore, there are 3*2*1 = 6 ways for exactly one of the toddler triplets to end up with their coat.

In other words, there are 6 possible scenarios in which exactly one of the toddler triplets get their coat.

For example, Red getting his coat, Green getting the wrong coat and Blue getting the wrong coat.

Answer 2

There are exactly 2 possible ways for one of the toddler triplets named Red, Green, and Blue to each get the correctly named coat when chosen at random. The solution involves considering specific scenarios rather than calculating permutations.

Step-by-step explanation:

The question asks for the number of ways for exactly one of the toddler triplets named Red, Green, and Blue to end up with the coat that matches their name if they choose at random.

This is not simply a permutation problem (n!), where all items are distinct and we're looking for all possible orders. To solve the problem, we have to consider the distinct scenarios that lead to exactly one child getting the correct coat. Let's consider the possible outcomes:

• Red picks his own coat (R), leaving Green and Blue with two choices. For Green to not get his coat (since we want only one child to get the correct coat), he must pick Blue's coat (B), leaving Green's coat (G) for Blue.
• Green picks his coat (G) first, Red would have to pick Blue's coat (B) next, leaving Red's coat (R) for Blue.
• Blue picks his coat (B) last, which means Green must have picked Red's coat (R), and Red must have picked Green's (G).

However, the third scenario doesn't work because Blue doesn't pick a coat at random but gets whatever is left. Therefore, we only have two scenarios--either Red picks his own, or Green picks his own, which means there are 2 possible ways for exactly one child to end up with the correct coat.

The correct answer is not 3! (6), as that would be the number of ways to arrange the coats if order mattered and all were choosing at random. In this problem, we're interested in a certain outcome, not all possible arrangements.