# Thread: Enumeration question

1. ## 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

2. 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.

. . $\displaystyle \_\;\square\;\_\;\square\;\_\;\square\;\_\;\square \;\_\;\square\;\_\;\square\;\_\;\square\;\_\;\squa re\;\_\;\square\:\_\;\square\;\_$

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

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

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

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.

. . $\displaystyle \_\;\square\;\_\;\square\;\_\;\square\;\_\; \square\;\_\;\square\;\_\;\square\;\_\;\square\;\_ \;\square\;\_$

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 $\displaystyle 3!= 6$ ways.

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

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

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 . . .

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

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