# combinations and permutations help

## combinations and permutations help

sorry for the abundance of math questions lately, this will probably be the last for now.

First read:

A 5 x 6 rectangle is formed using 1 x1 squares, 5 going up, 6 across. How many rectangles of various sizes can be found in this figure?

To solve:

[firstly, squares are a special type of rectangles]

Think of the grid as 7 vertical lines and 6 horizontal lines. To form a rectangle you must choose 2 vertical lines and 2 horizontal lines out of 6 possible. Each choice of 2 vertical and 2 horizontal gives a distinct rectangle.

So there are (7 choose 2) = 21 ways to choose the vertical lines

and (6 choose 2) = 15 ways to choose the horizontal lines

So there are 21*15 = 315 rectangles.

So this is done by the method of choosing, if you are familiar with it.

Now my actual question is, if instead it were a 5 x 5 square and it asked you to find how many

would you still be able to use

*just in case of you don't know, 6 choose 2, can be expressed as

First read:

A 5 x 6 rectangle is formed using 1 x1 squares, 5 going up, 6 across. How many rectangles of various sizes can be found in this figure?

To solve:

[firstly, squares are a special type of rectangles]

Think of the grid as 7 vertical lines and 6 horizontal lines. To form a rectangle you must choose 2 vertical lines and 2 horizontal lines out of 6 possible. Each choice of 2 vertical and 2 horizontal gives a distinct rectangle.

So there are (7 choose 2) = 21 ways to choose the vertical lines

and (6 choose 2) = 15 ways to choose the horizontal lines

So there are 21*15 = 315 rectangles.

So this is done by the method of choosing, if you are familiar with it.

Now my actual question is, if instead it were a 5 x 5 square and it asked you to find how many

*of various sizes that were in the 5 x 5 square,***squares**would you still be able to use

__choosing__to solve? And if yes, how?*just in case of you don't know, 6 choose 2, can be expressed as

**n! / (n-r)! r!**, or 6! / 4! x 2! = 15**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 23

Location : ubseikastan

## Re: combinations and permutations help

If a square within the 5x5 square is 5x5, there is only one square possible. If a square is 4x4, there are 4 different possible squares. The general pattern is (6 - length)

I don't see any

^{2}, because there are 6-length ways to move up and down (6-length unused rows blocks vertically when a length x length square is made), and there are 6-length ways to move horizontally. So, the answer is the sum of the series (6-length)^{2}from length=1 to length=5, which gives an answer of 55.I don't see any

*nice*way of doing this by choosing.**bfrsoccer**- Administrator
- Posts : 61

Join date : 2008-11-09

## Re: combinations and permutations help

yeah, your way was the same as the solution the paper had.

I was just wondering if like the first question, there was a fast effective way of using choosing to solve.

Thanks!

I was just wondering if like the first question, there was a fast effective way of using choosing to solve.

Thanks!

**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 23

Location : ubseikastan

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