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Math Help - Enumeration question

  1. #1
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    Question Enumeration question

    A store has 10 identical boxed stereos to store on 3 shelves. How many ways can the boxes be stored on the shelves if: (a) there are no restrictions on the number of stereos on each shelf? (b) the top shelf must have exactly 2 stereos on it? (c) each shelf must have at least 2 stereos on it? (d) no shelf can have more than 5 stereos on it? please help me thank you
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  2. #2
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    Hello, melody!

    A store has 10 identical boxed stereos to store on 3 shelves.
    How many ways can the boxes be stored on the shelves if:
    (a) There are no restrictions on the number of stereos on each shelf?

    Place the ten boxes in a row; insert a space before, after, and between them.

    . . \_\;\square\;\_\;\square\;\_\;\square\;\_\;\square  \;\_\;\square\;\_\;\square\;\_\;\square\;\_\;\squa  re\;\_\;\square\:\_\;\square\;\_<br />

    There are 11 spaces in which to insert 2 "dividers".


    If the two dividers are placed in two different spaces,
    . . there are: . {11\choose2} = 55 ways.
    For example: . \square\;\square\;\square\;\bigg| \;\square\;\square\;\square\;\square\;\square\;\bi  gg|\;\square\:\square \qquad\text{means: }\:\begin{Bmatrix} \text{3 on shelf 1} \\ \text{5 on shelf 2} \\ \text{2 on shelf 3}\end{Bmatrix}


    If the two dividers are placed in the same space,
    . . there are: . 11 ways.
    For example: . \square\;\square\;\square\;\square\;|\;|\;\square\  ;\square\;\square\;\square\;\square\:\square \qquad\text{ means: }\:\begin{Bmatrix}\text{4 on shelf 1} \\ \text{0 on shelf 2} \\ \text{6 on shelf 3} \end{Bmatrix}


    Hence, there are: .55 + 11 = 66 ways to place the boxes on the shelves.




    (b) The top shelf must have exactly 2 stereos on it.
    Place 2 boxes on the top shelf.
    Then the other 8 boxes are distributed to the other two shelves.

    Place the 8 boxes in a row; insert a space before, after, and between them.

    . . \_\;\square\;\_\;\square\;\_\;\square\;\_\; \square\;\_\;\square\;\_\;\square\;\_\;\square\;\_  \;\square\;\_

    There are 9 spaces in which to insert one divider.
    Hence, there are 9 ways to place the other 8 boxes.

    Therefore, there are 9 ways to have exactly two boxes on the top shelf.




    (c) Each shelf must have at least 2 stereos on it.
    For the last two parts, I had to make an exhaustive list.

    If a partition has three different numbers, for example, (1,3,6)
    . . the numbers can be assigned to the shelves in 3!= 6 ways.

    If a partition has two equal numbers, for example, (3,3,4)
    . . the numbers can be assinged to the shelves in 3 ways.

    . . \begin{array}{c|c}<br />
\text{Partition} & \text{Assign} \\ \hline<br />
(2,2,6) & 3 \\<br />
(2,3,5) & 6 \\<br />
(2,4,4) & 3 \\<br />
(3,3,4) & 3 \\ \hline<br />
\text{Total:} & 15 \end{array}


    There are 15 ways for each shelf to have at least two boxes.




    (d) No shelf can have more than 5 stereos on it.
    Another listing . . .

    . . \begin{array}{c|c}<br />
\text{Partition} & \text{Assign} \\ \hline<br />
(1,4,5) & 6 \\<br />
(2,3,5) & 6 \\<br />
(2,4,4) & 3 \\<br />
(3,3,4) & 3 \\ \hline<br />
\text{Total:} & 18 \end{array}

    There are 18 ways where each shelf has at most 5 boxes.

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