Probability, at least.. A florist sells 8 different types of flowers. Four people order one type of flower each.
What is the probability that at least 2 people order the same type?
I don't quite get this question.
I am thinking:
Number of ways to pick one type out of the eight = $_8C_1$
The total number of ways four people pick 2 of the same type = $_4C_2$
So the probability is $_4C_2/_8C_1$.
Just not confident that this is right.
 A: Always look at the context before substituting numbers in. So start with the basic equation:
$$p = {\text{# of cases where two pick the same type} \over \text{# of total cases}} = {\text{# of total cases} - \text{# of cases where all pick different types} \over \text{# of total cases}}$$
Now we have to find out the numbers to put in the numerator and denominator.
The number of total cases is $8^4$ because each person chooses independently of the others, so we simply raise the number of choices each person can make to the number of people.
The numerator is where permutations come in. The number of ways 4 people can each choose a different type is $P^8_4$, since we are permuting 8 types into 4 people with order significant.
So now we simply calculate:
$$p = {8^4 - P^8_4 \over 8^4} = {4096 - 1680 \over 4096} = \frac{2416}{4096} = \frac{151}{256}$$
A: Hint.  The answer is $1$ minus the probability that the four people all order different types of flowers.


*

*In how many ways can four people order four different types?  

*In how many ways can four people order four flowers altogether?


See how you go.
