# Complex Combination Problem

• Jul 20th 2008, 05:53 PM
raptorfactor
Complex Combination Problem
Hey guys... I've been doing logic questions all day and I'm stuck on this last one. Please lend me a hand and give me some help... :)

How many ways are there to pick a collection of 11 coins from piles of pennies, nickels, dime, and quarters? Base on the following condition:
a) Assuming that each pile has at least 11 coins.

I think this is....C(44,11)= 7669339132

b) There are only 10 coins in each pile

• Jul 23rd 2008, 02:17 PM
Plato
Quote:

Originally Posted by raptorfactor
How many ways are there to pick a collection of 11 coins from piles of pennies, nickels, dime, and quarters? Base on the following condition:
a) Assuming that each pile has at least 11 coins. [/FONT]

Basically we need to know how many solutions there are in the non-negative integers to the equation p+n+d+q=11, where the letters represent the numbers of coin types.
By common practice we regard the coins of any one type to be identical.

Quote:

Originally Posted by raptorfactor
How many ways are there to pick a collection of 11 coins from piles of pennies, nickels, dime, and quarters? Base on the following condition: b) There are only 10 coins in each pile

In this case, the only ways we found above that cannot be used are the 4 cases where all 11 coins are of the same type. So subtract.
• Jul 23rd 2008, 02:43 PM
abender
Hello Raptorfactor! I appreciate the hard work you have done today. Let's try to finish this up for ya!

Note that for combinations without repetitions, the number of combinations is given by the binomial coefficient, namely $(_{k}^n) = \frac{n!}{k!(n-k)!}$ where n is the number of objects you can pick from and k is the number of objects to be picked.

However, here we are dealing with combinations WITH repetitions. As such, the number of combinations is given by (n+k-1)! / (k!(n-1)!) where n is again the number of objects you can choose from and k is the number of objects to be chosen.

(Sorry, Latex is giving me a lot of trouble right now.)

A brief example: You go to McDonalds for a milkshake. Your parents tell you that you can get 2 milkshakes. McDonalds only sells vanilla, chocolate and strawberry (unless they came out with more within the past 15 years since I've been). Anyways, how many different combinations of milkshakes can you choose from? Use the formula above (the one not in latex)!

you want 2 milkshakes and choices are V C S

n = 3
k = 2

(3+2-1)! / (2!(3-1)!) = 4! / (2*2) = 4! / 4 = 3! = 6

Now double-checking:
V V
V C
V S
C C
C S
S S

Get part A and part B is just a little bit different.
Good luck!
-Andy