Imagine 4 coins, each coin has a heads and a tails, without knowing the math formula, you would draw a picture on paper of each possible orientation, you would come out with 16 different possibilities which is 4^2, (four squared).

Results 1 to 10 of 10

5*Thanks*

- March 9th 2014, 05:29 PM #1

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

- March 9th 2014, 07:24 PM #2

- Joined
- Feb 2014
- From
- Baguio, Philippines
- Posts
- 9
- Thanks
- 4

## Re: How Many Different Outfits

Imagine 4 coins, each coin has a heads and a tails, without knowing the math formula, you would draw a picture on paper of each possible orientation, you would come out with 16 different possibilities which is 4^2, (four squared).

- March 9th 2014, 08:38 PM #3

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

- March 10th 2014, 04:08 AM #4
## Re: How Many Different Outfits

@DuaneJack, The above answer is the wrong model for this question.

The correct answer is $2^4$. Although that gives the same numerical answer**the concept is different**.

Lets change the question by adding two different pairs of shoes.

**Now the answer is $2^5$ which is not the same as $5^2$.**

- March 10th 2014, 06:25 AM #5

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

- March 10th 2014, 06:27 AM #6

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

- March 10th 2014, 06:43 AM #7
## Re: How Many Different Outfits

There are no 'simple terms' here. From your other questions, I have concluded that you need more experience with the basics.

In basic terms, $2^4$ counts the number of functions from a set of four to a set of two; $4^2$ counts the number of functions from a set of two to a set of four. Now both equal 16, but clearly the two concepts are quite different.

$2^5$ counts the number of functions from a set of five to a set of two and $5^2$ counts the number of functions from a set of two to a set of five.

- March 10th 2014, 06:52 AM #8

- Joined
- Feb 2014
- From
- United States
- Posts
- 1,555
- Thanks
- 728

## Re: How Many Different Outfits

The point is this:

There are 2 possibilities for pants. For each of those, there are two possibilities for shirt, giving 4 = 2 * 2 possibilities for pants and shirt. For each of those, there are two possibilities for tie, giving 8 = 2 * 2 * 2 possibilities for pants, shirt, and tie. For each of those, there are two possibilities for jacket, giving

16 = 2 * 2 * 2 * 2 possibilities for pants, shoes, tie, and jacket. And if there were also two belts, there would be 32 = 2 * 2 * 2 * 2 * 2 possibilities.

The sequence is $2^n \ne n^2$ except by accident in the special case where n = 4 because $2^4 = 16 = 4^2.$

- March 10th 2014, 09:58 PM #9

- Joined
- Feb 2014
- From
- Baguio, Philippines
- Posts
- 9
- Thanks
- 4

## Re: How Many Different Outfits

And this special case is why you see it on tests, you know they love to throw you off when WE dont know the basics, as stated, thank you for the explaination, I have not seen a good basic explaination of this scenario, I still find myself drawing pictures of all permutations when the numbers are small, but now I think I have it so thanks.

If you have a deck of cards, 52, what are the chances of getting a 3 and a 7. Divide the cards into 4 groups for spades, clubs, hearts and diamonds, now there are only 13 cards to play chance, the answer is 2/13 but I dont know how to arrive at that answer logically. These simple ones make you feel stupid.

- March 11th 2014, 10:03 AM #10

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

## Re: How Many Different Outfits

I have been spending too much time with precalculus, calculus 1 and 2 material that easy math questions have become a bit fuzzy. I am learning calculus on my own while reviewing precalculus and geometry.

I love math.

I should have majored in math but selected another program. The cost of education today is ridiculous. I am learning calculus with the help of this great site and my calculus 1-3 dvds.

I plan to purchase the entire differential equations and linear algebra dvds from Jason Gibson. He is a fabulous tutor.

You can find his products at mathtutordvd.com.