# Counting Problems

1. Oct 21, 2004

### eku_girl83

Here are my problems:
1) Suppose seven school children (3 girls and 4 boys) are lined up to board a school bus. Find the number of ways they could line up if no two boys are together.
This arrangement would be BGBGBGB. So if I consider each BG as a unit, then there are 4! arrangements. But within the BG groupings, the B and G can switch places. So shouldn't the number of ways be 4!2!2!2! Is this correct?

2) Among ten lottery finalists, four will be selected to win individual amounts of $1000, 2000, 5000, and 10000. In how many ways can the money be distributed? I know that 10 choose 4 is 210. But don't I have to do something else? A permutation perhaps? Thanks for the help in advance! 2. Oct 21, 2004 ### NateTG 1. Because of the way the problem is stated, you can't have the boys and girls change places or you'd have two boys together. You also forgot that the girls can switch with each other. 2. Can one person win more than one amount of money? How many different people can win the$1000 prize ? After someone wins $1000, how many people can win$2000, after those two prizes have been awarded, how many different people can win $5000 ? After the other three prizes have been given out, how many choices are there for the$10000 winner?

3. Oct 22, 2004

### robert Ihnot

BGBGBGB is the only possible way to seperate them. Send all 4 boys to one spot and all three girls to another. Then the teacher can fill the arranged slots by chosing from the boys in 4! ways and from the girls in 3! ways, so that is 144 ways.

In the second problem there are 10C4 = 210 ways of chosing 4 from 10, but those four can be arranged in 4! ways, so that is 5040 ways.

Or if you'd rather just consider it a permutation problem, there are 10 ways to chose the first prize person, 9 for the second, up to the 4th. So that is the permutation:

$$\frac{10!}{6!}=5040$$

Last edited: Oct 22, 2004