2013 AMC 12A Problems/Problem 15

Problem

Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?

$\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372$

Solution 1

There are two possibilities regarding the parents.

1) Both are in the same store. In this case, we can treat them both as a single bunny, and they can go in any of the 4 stores. The 3 baby bunnies can go in any of the remaining 3 stores. There are $4 \cdot 3^3 = 108$ combinations.

2) The two are in different stores. In this case, one can go in any of the 4 stores, and the other can go in any of the 3 remaining stores. The 3 baby bunnies can each go in any of the remaining 2 stores. There are $4 \cdot 3 \cdot 2^3 = 96$ combinations.

Adding up, we get $108 + 96 = 204$ combinations.

Solution 2

We tackle the problem by sorting it by how many stores are involved in the transaction.

1) $2$ stores are involved. There are $\binom{4}{2} = 6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6 \cdot 2 = 12$ total arrangements.

2) $3$ stores are involved. There are $\binom{4}{3} = 4$ ways to choose which of the stores are involved. We then break the problem down to into two subsections - when the parents and grouped together or sold separately.

Separately: All children must be in one store. There are $3!$ ways to arrange this. $6$ ways in total.

Together: Both parents are in one store and the 3 children are split between the other two. There are $\binom{3}{2}$ ways to split the children and $3!$ ways to choose to which store each group will be sold. $3! \cdot \binom{3}{2} = 18$ .

$(6 + 18) \cdot 4 = 96$ total arrangements.

3) All $4$ stores are involved. We break down the problem as previously shown.

Separately: All children must be split between two stores. There are $\binom{3}{2} = 3$ ways to arrange this. We can then arrange which group is sold to which store in $4!$ ways. $4! \cdot 3 = 72$ .

Together: Both parents are in one store and the 3 children are each in another store. There are $4! = 24$ ways to arrange this.

$24 + 72 = 96$ total arrangements.

Final Answer: $12 + 96 + 96 = \boxed{\textbf{(D)} \: 204}$ .

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2013 AMC 12A Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See also