Four friends, standing in a row for a picture, are in the following order:

## Anyu Brad Chi Diego

The friends then move so that Anyu is not in the $1^{\text {st }}$ position, Brad is not in the $2^{\text {nd }}$ position, Chi is not in the $3^{\text {rd }}$ position, and Diego is not in the $4^{\text {th }}$ position. In how many ways can the friends do this? That is, in how many ways can they rearrange themselves so that each person is not in their original position?
(A) 6
(B) 9
(C) 12
(D) 18
(E) 24

## Part C: Each correct answer is worth 8.
Let Anyu, Brad, Chi, and Diego be represented by $A, B, C$, and $D$, respectively, and so their original order is $A B C D$.

When rearranged, $A$ is not in the $1^{\text {st }}$ position, and so there are exactly 3 cases to consider: $A$ is in the $2^{\text {nd }}$ position, or $A$ is in the $3^{\text {rd }}$ position, or $A$ is in the $4^{\text {th }}$ position.

For each of these 3 cases, we count the number of ways to arrange $B, C$, and $D$.

Case 1: $A$ is in the $2^{\text {nd }}$ position

Since $A$ is in the $2^{\text {nd }}$ position, $B$ can be in any of the other 3 positions $\left(1^{\text {st }}, 3^{\text {rd }}\right.$ or $\left.4^{\text {th }}\right)$.

If $B$ is in the $1^{\text {st }}$ position, then there is exactly one possible rearrangement: $B A D C$ (since $C$ and $D$ cannot be in the $3^{\text {rd }}$ and $4^{\text {th }}$ positions respectively).

If $B$ is in the $3^{\text {rd }}$ position, then there is exactly one possible rearrangement: $D A B C$ (since $D$ cannot be in the $4^{\text {th }}$ position).

If $B$ is in the $4^{\text {th }}$ position, then there is exactly one possible rearrangement: $C A D B$ (since $C$ cannot be in the $3^{\text {rd }}$ position).

Thus there are exactly 3 possible rearrangements when $A$ is in the $2^{\text {nd }}$ position.

Case 2: $A$ is in the $3^{\text {rd }}$ position

Since $A$ is in the $3^{\text {rd }}$ position, $C$ can be in any of the other 3 positions.

In a manner similar to Case 1 , it can be shown that there are 3 possible rearrangements in this case: $C D A B, D C A B$, and $B D A C$.

Case 3: $A$ is in the $4^{\text {th }}$ position

Since $A$ is in the $4^{\text {th }}$ position, $D$ can be in any of the other 3 positions.

Similarly, there are 3 possible rearrangements in this case: $D C B A, C D B A$, and $B C D A$.

So that each person is not in their original position, the four friends can rearrange themselves in $3+3+3=9$ different ways.

(Such a rearrangement of a list in which no element appears in its original position is called a derangement.)

ANswer: (B)