# Counting

• May 8th 2010, 09:24 PM
chubbs145
Counting
In how many ways can 6 different candies be passed out to nine people if everyone can receive just one?

can we just assume that they have a choice of 6 each time? Like:
6*6*6*6*6*6*6*6*6

The next problem asks the same question but in this case everyone can have any number of candies....if the first part isn't right then this part won't be right.

Any clarification?
• May 8th 2010, 10:16 PM
slovakiamaths
Quote:

Originally Posted by chubbs145
In how many ways can 6 different candies be passed out to nine people if everyone can receive just one?

can we just assume that they have a choice of 6 each time? Like:
6*6*6*6*6*6*6*6*6

The next problem asks the same question but in this case everyone can have any number of candies....if the first part isn't right then this part won't be right.

Any clarification?

$^9c_6=\frac{9*8*7}{3*2*1}=84$
• May 9th 2010, 05:14 AM
Plato
Quote:

Originally Posted by chubbs145
In how many ways can 6 different candies be passed out to nine people if everyone can receive just one?

This is a simple permutation: $^9\mathcal{P}_6=\frac{9!}{(9-6)!}=9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4~$.
• May 9th 2010, 05:34 AM
Soroban
Hello, chubbs145!

Quote:

In how many ways can 6 different candies be passed out to nine people
if everyone can receive just one?

Plato is absolutely correct!

Quote:

The next problem asks the same question
but in this case, everyone can have any number of candies.

For each of the 6 candies, there are 9 choices of people to give it to.

Answer: . $9^6 \:=\:531,\!441$ ways.
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