7. In how many different ways can all the letters of the word ANAGRAM be arranged so that no two $A$s are ever next to each other?

The use of a pocket calculator or any reference materials is not allowed.
First solution.

We will specifically determine the number of ways to arrange three letters $A$ in some of the 7 positions and the number of ways to arrange the remaining letters $N, G, R, M$ in the remaining 4 positions. Then we will multiply these numbers.

Determine the number of ways to arrange three letters $A$.

If the first letter $A$ is in the first position, then the second letter $A$ can be in:

- the third position, and the third letter $A$ in the fifth, sixth, or seventh position (3 ways)
- the fourth position, and the third letter $A$ in the sixth or seventh position (2 ways)
- the fifth position, and the third letter $A$ in the seventh position (1 way)

1 POINT

If the first letter $A$ is in the second position, then the second letter $A$ can be in:

- the fourth position, and the third letter $A$ in the sixth or seventh position (2 ways)
- the fifth position, and the third letter $A$ in the seventh position (1 way)

1 POINT

If the first letter $A$ is in the third position, then the second letter $A$ can be in:

- the fifth position, and the third letter $A$ in the seventh position (1 way)

1 POINT

We have a total of 10 such ways to arrange the three letters $A$.

The previous counting procedure and corresponding scoring are also described in the table.

Let $* \in\{N, G, R, M\}$.

| $A * A * A * *$ | $A * * A * A *$ | $A * * * A * A$ | $* A * A * A *$ | $* A * * A * A$ | $* * A * A * A$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $A * A * * A *$ | $A * * A * * A$ |  | $* A * A * * A$ |  |  |
| $A * A * * * A$ |  |  | 1 POINT |  | 1 POINT |
| 2 POINTS |  |  |  |  |  |

Determine the number of ways to arrange the letters $N, G, R, M$ in the remaining 4 positions.

They can be arranged in $4 \cdot 3 \cdot 2 \cdot 1=24$ ways.

3 POINTS

Finally, all the letters of the word ANAGRAM can be arranged in 240 different ways.

1 POINT TOTAL 10 POINTS

Note 1: If a student first writes all the possibilities where $A$ is in the first position and gets 6 solutions and concludes that symmetrically (when $A$ is in the last position) they will get another 6 solutions, and then subtracts 3 (since that is the number of possibilities where $A$ is in both the first and last positions), they will incorrectly conclude that the answer is 9. Indeed, they have missed the possibility that $A$ is neither in the first nor in the last position. Then the student should get 3 POINTS out of the 4 POINTS allocated for the number of ways to arrange the letters $A$.

Note 2: The last 1 POINT is awarded to the student only with the previous argumentation.