# Math Help - Combinations and a bit of probability

1. ## Combinations and a bit of probability

Another set of interesting questions:

1. In how many ways can four squares, not all in the same row or column be selected from an 8-by-8 chessboard to form a rectangle?

2. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Find the probability that no two birch trees are next to one another.

3. 25 people sit around a circular table. Three of them are chosen randomly. What is the probability that at least two of the three are sitting next to one another?

2. Originally Posted by usagi_killer
1. In how many ways can four squares, not all in the same row or column be selected from an 8-by-8 chessboard to form a rectangle?
${8\choose2} = 28$ ways of choosing which two rows, same number of ways of choosing which two columns, total of $28^2 = 784$ rectangles.

That assumes that the rectangles have sides parallel to the sides of the board. If you are not making that assumption then there will be many other possibilities, such as when the four squares are a knight's move from their neighbours.

3. 2. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Find the probability that no two birch trees are next to one another.
Here is how I would approach this one. There are always different approaches to counting problems.

Arrange the Maples and Oaks in 7!=5040 ways. This now gives us 8 spaces in between to put the birch trees so they are separated and not next to one another. We can choose these spots in
C(8,5)=56 ways and 5!=120 ways to arrange them.

There is 12! ways to arrange them without restriction.

So, the probability of no birch being next to another is

$\frac{\binom{8}{5}\cdot 5!\cdot 7!}{12!}=\frac{7}{99}$