1. ## Small problem

Five teachers at a meeting remove their name tags and put them in a bag. One by one, each teacher randomly draws one tag from the bag until the bag is empty. What is the probability that exactly one of the teachers draws the correct name tag?

i got this so far...

one person gets their name and four of the people dont get their names: P(that)= 1/5 * 3/4 * 2/3 * 1/2 * 1 = .05

then since the order may happen any given way, we multiply by the number of arrangements possible: 5 "5 choose 1 = 5!/[4!1!]"

ans = .25

2. Originally Posted by ruthvik
[FONT=arial]Five teachers at a meeting remove their name tags and put them in a bag. One by one, each teacher randomly draws one tag from the bag until the bag is empty. What is the probability that exactly one of the teachers draws the correct name tag?
There are 120 ways for the name tags to endup.
There are 44 ways in which no one gets his/her own name tag.
How many get ways are there for someone to get the correct name tag?
There are 5 ways for exactly one to be correct and 9 ways for the orhers to be incorrect.

3. Hello, ruthvik!

This is not a simple problem . . .

Five teachers at a meeting remove their name tags and put them in a bag.
One by one, each teacher randomly draws one tag from the bag until the bag is empty.
What is the probability that exactly one of the teachers draws the correct name tag?

As Plato pointed out, there are: . $5! \:=\:120$ ways the tags could be distributed.

Now, exactly one person gets his/her tage.
. . There are ${\color{red}5}$ choices for that person.

Then the other four must not get their own tags.

Suppose their names are $A,B,C,D.$

$\begin{array}{cc}\text{The tags could given like this:} \\ \\[-4mm]
BADC \\ BCDA \\ BDAC \\ CADB \\ CDAB & {\color{red}9}\text{ ways}\\ CDBA \\ DABC \\ DCAB \\ DCBA \end{array}$

Hence, there are: . $5\cdot9 \:=\:45$ ways.

The probability is: . $\frac{45}{120} \;=\;\frac{3}{8}$