# Math Help - combinations permutations counting and probability?

1. ## combinations permutations counting and probability?

It turns out counting things really isn't that easy. Any help with explanations and formulae would be greatly appreciated.

A group of 30 people consists of 15 men and 15 women. Each of the 30 choose one cooldrink from 10 types:
(1) How many ways could they do this? (I'm thinking 10^30?)
(2) How many different combinations of cooldrinks could be ordered from the cafeteria where you don't care who gets which but only want to get the number of each type right? (is this in terms of n because they don't tell us the number of each type? Formula?I don't understand this question)

Given the group orders
15 of type 1
6 of type 2
3 of type 3
6 of type 4
(3)In how many different ways can I distribute the cooldrinks amongst the 30 people if I don't care who ordered what?(30! ways?)
(4) What is the probability that a random division of the cooldrink amongst the 30 people give each what they ordered?

I don't even no where to start with the second and fourth questions so any input would be an immense help. Thanks in advance

2. Originally Posted by chocaholic
A group of 30 people consists of 15 men and 15 women. Each of the 30 choose one cooldrink from 10 types:
(1) How many ways could they do this? (I'm thinking 10^30?)
(2) How many different combinations of cooldrinks could be ordered from the cafeteria where you don't care who gets which but only want to get the number of each type right? (is this in terms of n because they don't tell us the number of each type? Formula?I don't understand this question)
Given the group orders
15 of type 1
6 of type 2
3 of type 3
6 of type 4
(3)In how many different ways can I distribute the cooldrinks amongst the 30 people if I don't care who ordered what?(30! ways?)
(4) What is the probability that a random division of the cooldrink amongst the 30 people give each what they ordered?
Number 1 is correct.
Number 2 is a multi-selection (multi-set) problem.
The number of ways to place $N$ identical objects into $K$ distinct cells is $\binom{N+K-1}{N}$.
So in #2 $N=30~\&~K=10$, we are selecting 30 from 10.

#3 is a matter of combinations: $\binom{30}{15}\binom{15}{6} \binom{9}{3} \binom{6}{6}$

What do you think #3 has to do with #4?

3. Hi Plato

Thanks for the input.I still don't quite understand the multi-set one but I'll read up on it. As to the relation between #3 and #4 I think the probability=the number of combinations /the ways to distribute =answer#2/answer#3 Am I on the right track?