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Math Help - Combinatory

  1. #1
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    Combinatory

    I am having trouble solving for this problem. I would like to see a clear way to obtain the solution.
    How many ways you can distribute 10 different toys to 4 children if each child receives at least 2 toys?
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  2. #2
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    Re: Combinatory

    What are the possible outcomes? One child can receive 4 toys, each other gets 2. Two children can receive 3 toys, the other two get 2. We will handle each case separately.

    For the case that one child receives 4 toys and the other two get 2 each, choose the child who receives 4 toys. There are 4 children to choose and you choose 1 who will receive the 4 toys. Then, choose 4 of the 10 toys to give to the first child. Then, choose 2 of the remaining 6 toys to give to the second child, 2 of the remaining 4 to give to the third child, and 2 of the remaining two to give to the last child.

    So, for this case, there are
    $\underbrace{\binom{4}{1}}_{\begin{matrix}\text{Ch oose kid} \\ \text{to get 4 toys}\end{matrix}} \underbrace{\binom{10}{4}}_{\begin{matrix}\text{Ch oose 4 toys} \\ \text{for first child}\end{matrix}} \underbrace{\binom{6}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for second child}\end{matrix}} \underbrace{\binom{4}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for third child}\end{matrix}} \underbrace{\binom{2}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for last child}\end{matrix}}$

    This can be simplified as:

    $\dfrac{4!\cdot 10!}{(1!3!)(4!2!2!2!)} = 75,600$

    Next, choose two children to get 3 toys. Then choose 3 of the 10 toys for the first child, 3 of the remaining 7 for the second child, 2 of the remaining 4 for the third, and 2 of the remaining 2 for the last:

    $\underbrace{\binom{4}{2}}_{\begin{matrix}\text{Ch oose kids} \\ \text{to get 3 toys}\end{matrix}} \underbrace{\binom{10}{3}}_{\begin{matrix}\text{Ch oose 3 toys} \\ \text{for first child}\end{matrix}} \underbrace{\binom{7}{3}}_{\begin{matrix}\text{Cho ose 3 toys} \\ \text{for second child}\end{matrix}} \underbrace{\binom{4}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for third child}\end{matrix}} \underbrace{\binom{2}{2}}_{\begin{matrix}\text{Cho ose 2 toys} \\ \text{for last child}\end{matrix}}$

    This can be simplified as:

    $\dfrac{4!\cdot 10!}{(2!2!)(3!3!2!2!)} = 151,200$

    Adding these together: $75,600+151,200 = 226,800$
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  3. #3
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    Re: Combinatory

    Hello, Yeison!

    SlipEternal is absolutely correct.
    Here is another approach.


    How many ways you can distribute 10 different toys to 4 children
    if each child receives at least 2 toys?

    There are two possible distributions: . \begin{Bmatrix}[1] & (2,2,2,4)  \\ [2] & (2,2,3,3)\end{Bmatrix}


    Case [1]

    There are 4 choices for the child to receive four toys.
    Then the toys can be distributed in: {10\choose2,2,2,4} = 18,\!900 ways.
    . . There are: . 4\cdot18,\!900 \,=\,75,\!600 ways.


    Case [2]
    There are {4\choose2,2} = 6 choices for the children to receive three toys.
    Then the toys can be distributed in: {10\choose2,2,3,3} = 25,\!200 ways.
    . . There are: . 6\cdot 25,\!200 \:=\:151,\!200 ways.


    Therefore, there are . 75,\!600 + 151,\!200 \:=\:226,\!800 ways.
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  4. #4
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    Re: Combinatory

    Thank you. This explanation has been real helpful.
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