### Instruction: How many ways are there to arrange the letters of the word $\text{BA}_1\text{N}_1\text{A}_2\text{N}_2\text{A}_3$, in which the three A's and the two N's are considered different?### Response: To arrange the letters of the word, I need to assign each letter a position from 1 to 6.
There are 6 positions to choose from for the B, so there are 6 ways to place the B.
After placing the B, there are 5 positions left for the $\text{A}_1$, so there are 6 times 5 ways to place the B and the $\text{A}_1$.
Similarly, after placing the B and the $\text{A}_1$, there are 4 positions left for the $\text{N}_1$, so there are 6 times 5 times 4 ways to place the B, the $\text{A}_1$, and the $\text{N}_1$.
Continuing this way, after placing the B, the $\text{A}_1$, and the $\text{N}_1$, there are 3 positions left for the $\text{A}_2$, so there are 6 times 5 times 4 times 3 ways to place the B, the $\text{A}_1$, the $\text{N}_1$, and the $\text{A}_2$.
Then, there are 2 positions left for the $\text{N}_2$, so there are 6 times 5 times 4 times 3 times 2 ways to place the B, the $\text{A}_1$, the $\text{N}_1$, the $\text{A}_2$, and the $\text{N}_2$.
Finally, there is only 1 position left for the $\text{A}_3$, so there are 6 times 5 times 4 times 3 times 2 times 1 ways to place all the letters.
This is the same as calculating 6 factorial, or 6!, which is equal to 720.