I've been struggling with these word problems, I feel like they should be easy enough but I keep getting the wrong answer. So I'm probably approaching them wrong

1. In how many ways can four colas, three iced teas, and three orange juices be distributed to ten graduates if each grad is to receive one beverage?

2. Assuming that at least one coin is used, how many different sums of money, can be made from the following coins: a penny, a nickel, a dime, a quarter and a dollar coin?

80 fg to whoever can solve those 2 problems with an explanation and this one

7Pr = 42

7!/(7-r)! = 42

I have the answers, I just don't know how to get to them.

This post was edited by Jwoww on Jan 13 2014 03:41am

not sure what 7Pr = 42 is supposed to mean

7!/(7-r)! = 42 is easy with 42/7=6

7*6*(5*4*3*2*1)/(7-r)!=42 requires 7-r to be 5, ie r=2

Problem 2 :

You have 32 possible choices :
0 or 1 penny -> 2 choices
0 or 1 nickel -> 2 choices
0 or 1 dime -> 2 choices
0 or 1 quarter -> 2 choices
0 or 1 dollar -> 2 choices

2*2*2*2*2 = 2^5 = 32

You must substract 1 because you're not allowed to select no coin at all : 32 - 1 = 31.

It's easy to check that those 31 choices all lead to different sums of money.

Problem 1 :

The graduates with colas form a 4-combination of the set of the 10 graduates.
There are ( 10 4 ) = 10*9*8*7 / ( 4*3*2*1) such combinations.

Once the colas have been taken, the graduates with iced teas form a 3-combination of the set of the 6 remaining graduates.
There are ( 6 3 ) = 6*5*4 / (3*2*1) such combinations.

And the last graduates take orange juices.

Finally : ( 10 4 ) * ( 6 3 ) * 1 = 4,200 ways.

Thank you guys, wrote the test today and it wasn't bad at all. Since both of you contributed I split the 80fg amongst both of you.

Thank you very much