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Math Help - Counting/Combinations

  1. #1
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    Counting/Combinations

    I have two questions regarding counting and permutations that I am struggling with. Some help would be appreciated.

    1) There is a bike race with 1001 participants, 670 woman and 331 men. The first five participants to cross the finish line are awarded a medal (gold, silver, bronze, copper, and iron), and the last one gets a pink bike horn. Additionally, all participants get a t-shirt, but the organizers only have 500 t-shirts to give out, so they decide to give them to the first 500 participants to finish the race.

    i) Determine the number of possible ways that all the awards (medals, t-shirts, and bike horn) can be distributed. (Assume all t-shirts are identical and that everyone finishes the race).

    My solution:

    The number of ways that the 6 awards can be distributed is  \frac{1001!}{(1001-6)!} = 9.91024985^{17}
    The number of the t shirts can be distributed is  \binom{495}{495} = 1 ( since the 5 medal winners also got t shirts). So this doesn't have any effect of the result above.

    I believe that this part is correct but please correct me if I'm wrong.

    ii)Count the number of possible outcomes of the race in which exactly three women get a medal and exactly 370 woman get a t shirt.

    My solution (attempt):

    The top 5 participants have to have 3 woman and 2 men, and can be represented by  WWWMM
    so the number of ways this can be arranged is \frac{5!}{3!2!} = 10
    and the number of ways that 670 woman can be places in these 3 positions is  \frac{670!}{667!}=299417640 .

    Now for the t shirts, I'm not sure what to do. I tried doing \binom{495}{367} since there are 495 t shirts left and 267 woman still need a t shirt. I don't think this is correct though since it was too large for my calculator to compute. Also, do I add this number to the number of ways the medals can be distributed or multiply it?


    2)63 people are going on a camping trip. They have two 6-person tents, three 4-person tents, five 3-person tents, and three 2-person tents. 18 people sleep outside of the tents under a giant tarp.

    i)Determine in how many ways the students can choose the tents to sleep. (Assume the tents are different, regardless of size.)

    My solution attempt:

    First, 18 people out of 63 are chosen for the tarp, then 6 people out of the 45 left are chosen for the 6 person tent, then 6 more out of the 39 left are chosen for the next 6 person tent, and so on until there are no more people left.

    So the number of ways people can choose to sleep in the tents is  \frac{\binom{63}{18} \binom{45}{6} \binom{39}{6} ... \binom{2}{2}}{14!} = 1.06 *10^{46}

    ii)Suppose that there are 4 people that snore. Count the number of ways of assigning the people to tents so that all snorers share their tents only with other snorers.


    I have no idea where to even begin with this one, some help would be appreciated.

    Thanks in advance.
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  2. #2
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    Re: Counting/Combinations

    Did I post this in the wrong place or something?
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  3. #3
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    Re: Counting/Combinations

    Hello, nubshat!

    2) 63 people are going on a camping trip. They have two 6-person tents, three 4-person tents,
    five 3-person tents, and three 2-person tents. 18 people sleep outside of the tents under a giant tarp.

    1) Determine in how many ways the students can choose the tents to sleep.
    (Assume the tents are different, regardless of size.)

    This is a partition: . \frac{63!}{(6!)^2(4!)^3(3!)^5(2!)^318!}




    2) Suppose that there are 4 people that snore. Count the number of ways of assigning
    the people to tents so that all snorers share their tents only with other snorers.

    Assign the 4 snorers to one of the 4-person tents.
    There are 3 choices.

    Then the other 59 people will be assigned in . \frac{59!}{(6!)^2(4!)^2(3!)^5(2!)^318!} ways.

    Answer: . 3\cdot\frac{59!}{(6!)^2(4!)^2(3!)^5(2!)^318!} ways.
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  4. #4
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    Re: Counting/Combinations

    Thanks for the reply, just one question about part 2, isn't another option to put the 4 snorers in two 2-person tents? which would create 6 more ways of arranging the campers?
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  5. #5
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    Re: Counting/Combinations

    Hello again, nubshat!

    isn't another option to put the 4 snorers in two 2-person tents?

    Absolutely right!
    I completely overlooked that situation . . . *blush*


    First, select two of the three 2-person tents: . 3 choices.

    Then assign the 4 snorers to the two tents: . {4\choose2,2} ways.

    Then assign tents to the other 59 people: . \;\frac{59!}{(6!)^2(4!)^3(3!)^5(2!)(18!)} ways.

    Multiply.

    Thanks from nubshat
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  6. #6
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    Re: Counting/Combinations

    You need to revisit your first problem. Let's approach it a different way.

    How many ways can we choose 500 winners out of 1001 racers. $\dbinom{1001}{500} = \dfrac{1000!}{500! *501!}.$

    Out of the 501 racers who came in last, how many ways can we select the one who came in dead last?

    Of the 500 racers who came in first, how many ways can we choose numbers 1 through 5?
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  7. #7
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    Re: Counting/Combinations

    Quote Originally Posted by JeffM View Post
    You need to revisit your first problem. Let's approach it a different way.

    How many ways can we choose 500 winners out of 1001 racers. $\dbinom{1001}{500} = \dfrac{1000!}{500! *501!}.$

    Out of the 501 racers who came in last, how many ways can we select the one who came in dead last?

    Of the 500 racers who came in first, how many ways can we choose numbers 1 through 5?
    Like you said, there are \frac{1001!}{500!501!} ways to choose 500 winners.
    There are \frac{501!}{500!} ways to choose the last place
    and \frac{500!}{495!} ways to choose the medal winners.

    So there are  \frac{1001!}{500!501!} * \frac{501!}{500!} * \frac{500!}{495!} ways to distribute all the awards. Is this correct? Or am I missing something?
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  8. #8
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    Re: Counting/Combinations

    The only thing I would say in addition is that you can simplify that expression.
    Thanks from nubshat
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