Results 1 to 2 of 2

Math Help - Choosing a committee and distributing toppings on pizza.

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    83

    Choosing a committee and distributing toppings on pizza.

    Hey! I am having trouble with a question. I believe I have found the solution to the first two portions of the question, but I am having trouble with the last bit. I am going to post the entire question, as well as my solution, and hope that someone here can help me out!

    Question
    The student council is ordering pizza for their next meeting. There are 20 council members, 7 of whom are vegetarian. A committee of 3 will order 6 pizzas from a pizza shop that has a special price for large pizzas with up to three toppings. The shop offers 10 different toppings.
    a) How many different pizza committees can the council choose if there must be at least 1 vegetarian and 1 non-vegetarian on the committee?
    b) In how many ways could the committee choose up to 3 toppings for a pizza?
    c) The committee wants as much variety as possible in the toppings. They decide to order each toping exactly once and to have at least 1 topping on each pizza. Describe the different cases possible when distributing the toppings in this way.
    d) For one of these cases, determine the number of ways of coosing and distributing the 10 toppings.

    Solution
    a) There are two possible variations in the committee:
    1 vegetarian: (7C1)(13C2) = 546
    2 vegetarian: (7C2)(13C1) = 273
    546+273 = 819 different pizza committees possible.

    b) 0 toppings = (10C0) = 1
    1 topping = 10C1 = 10
    2 toppings = 10C2 = 45
    3 toppings = 10C3 = 120
    1 + 10 + 45 + 120 = 176 ways to choose up to 3 toppings on a pizza.

    c) So each pizza needs to have at least one topping, and they are only ordering each topping once. So I know one possibility is 1 topping on 4 of the pizzas and 2 on 3 of them. Another is 3 on one pizza and 2 on another and 1 on each of the other 5. This is all of the combinations I can think of. When you read the question, is this how you thought of answering it? I am rather confused how to go about finding the solution for this.

    d) Okay, I am completely lost in how to find the solution to this problem. This and part c were the main reasons for posting this question. Please help!

    Hopefully my solutions make sense. If you can help me in any way shape or form it is greatly appreciated!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Kakariki View Post
    Hey! I am having trouble with a question. I believe I have found the solution to the first two portions of the question, but I am having trouble with the last bit. I am going to post the entire question, as well as my solution, and hope that someone here can help me out!

    Question
    The student council is ordering pizza for their next meeting. There are 20 council members, 7 of whom are vegetarian. A committee of 3 will order 6 pizzas from a pizza shop that has a special price for large pizzas with up to three toppings. The shop offers 10 different toppings.
    a) How many different pizza committees can the council choose if there must be at least 1 vegetarian and 1 non-vegetarian on the committee?
    b) In how many ways could the committee choose up to 3 toppings for a pizza?
    c) The committee wants as much variety as possible in the toppings. They decide to order each toping exactly once and to have at least 1 topping on each pizza. Describe the different cases possible when distributing the toppings in this way.
    d) For one of these cases, determine the number of ways of coosing and distributing the 10 toppings.

    Solution
    a) There are two possible variations in the committee:
    1 vegetarian: (7C1)(13C2) = 546
    2 vegetarian: (7C2)(13C1) = 273
    546+273 = 819 different pizza committees possible.

    b) 0 toppings = (10C0) = 1
    1 topping = 10C1 = 10
    2 toppings = 10C2 = 45
    3 toppings = 10C3 = 120
    1 + 10 + 45 + 120 = 176 ways to choose up to 3 toppings on a pizza.

    c) So each pizza needs to have at least one topping, and they are only ordering each topping once. So I know one possibility is 1 topping on 4 of the pizzas and 2 on 3 of them. Another is 3 on one pizza and 2 on another and 1 on each of the other 5. This is all of the combinations I can think of. When you read the question, is this how you thought of answering it? I am rather confused how to go about finding the solution for this.

    d) Okay, I am completely lost in how to find the solution to this problem. This and part c were the main reasons for posting this question. Please help!

    Hopefully my solutions make sense. If you can help me in any way shape or form it is greatly appreciated!!!
    a) Agreed

    b) Agreed

    c) This would be correct if the committee were ordering 7 pizzas -- but they're only ordering 6 of them! So we have

    {3,3,1,1,1,1}
    {3,2,2,1,1,1}
    {2,2,2,2,1,1}

    which is 3 cases. I listed them methodically to ensure there are no missing cases.

    d) I'll use C(n,k) where you wrote nCk. Consider case

    {3,3,1,1,1,1}

    There are C(10,4) ways to choose {1,1,1,1}. Then there are C(6,3) ways to choose one of the {3}, which completely determines the other {3}. But we count each of these twice, for example {mushrooms,pepperoni,green peppers},{pineapple,ham,spinach} is counted as well as {pineapple,ham,spinach},{mushrooms,pepperoni,green peppers}. So for this case it's (1/2) * C(10,4) * C(6,3).

    The second case is similar, I'll let you do that one.

    The last case is a bit different, in terms of the {2,2,2,2} part. So we take care of {1,1} and now have 8 toppings left, I'll call them {a,b,c,d,e,f,g,h}, and label the pizzas {p1,p2,p3,p4}. Let p1 be the pizza with topping "a". There are 7 ways to make this pizza. Now there are 6 toppings left, relabel them {k,l,m,n,o,p} and let p2 be the pizza with topping "k." There are 5 ways to choose p2. Relabel again and similarly there are 3 ways to choose p3, which determines p4. So altogether we get C(10,2) * 7 * 5 * 3.

    There might be a better way to do that last case.

    P.S. This is my 1000th post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. choosing a committee
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 12th 2010, 12:01 PM
  2. choosing pizza toppings
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: May 11th 2010, 07:42 PM
  3. whos eaten more pizza?
    Posted in the Algebra Forum
    Replies: 11
    Last Post: December 2nd 2009, 02:17 AM
  4. Combinations - Pizza Toppings
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 6th 2009, 06:43 AM
  5. Time for Pizza
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 2nd 2008, 04:49 AM

Search Tags


/mathhelpforum @mathhelpforum