Results 1 to 10 of 10
Like Tree5Thanks
  • 1 Post By DuaneJack
  • 1 Post By Plato
  • 1 Post By Plato
  • 1 Post By JeffM
  • 1 Post By DuaneJack

Math Help - How Many Different Outfits

  1. #1
    Banned
    Joined
    Jan 2013
    From
    New York City
    Posts
    658
    Thanks
    7

    How Many Different Outfits

    A man packed 2 pants, 2 shirts, 2 jackets, and 2 ties. How many different outfits can he wear that includes one of each item?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    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).
    Thanks from nycmath
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Jan 2013
    From
    New York City
    Posts
    658
    Thanks
    7

    Re: How Many Different Outfits

    Easily stated.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,959
    Thanks
    1783
    Awards
    1

    Re: How Many Different Outfits

    Quote Originally Posted by nycmath View Post
    A man packed 2 pants, 2 shirts, 2 jackets, and 2 ties. How many different outfits can he wear that includes one of each item?
    Quote Originally Posted by DuaneJack View Post
    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).
    @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 $5^2$.
    Thanks from nycmath
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Jan 2013
    From
    New York City
    Posts
    658
    Thanks
    7

    Re: How Many Different Outfits

    Thank you very much, Plato.
    Thank you DuaneJack for sure effort.
    Probability word problems are fuzzy.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Jan 2013
    From
    New York City
    Posts
    658
    Thanks
    7

    Re: How Many Different Outfits

    Quote Originally Posted by Plato View Post
    @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 $5^2$.
    Can you explain in simple terms why the concept is different?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,959
    Thanks
    1783
    Awards
    1

    Re: How Many Different Outfits

    Quote Originally Posted by nycmath View Post
    Can you explain in simple terms why the concept is different?
    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.
    Thanks from nycmath
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Feb 2014
    From
    United States
    Posts
    817
    Thanks
    410

    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 $2^4 = 16 = 4^2.$
    Thanks from nycmath
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    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.
    Thanks from nycmath
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Jan 2013
    From
    New York City
    Posts
    658
    Thanks
    7

    Re: How Many Different Outfits

    Quote Originally Posted by DuaneJack View Post
    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.

    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.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum