I wanted to ask if I got the answers to the following two (homework) questions right, and if not whether someone could give me a hint where I went wrong?
We choose $10$ numbers with repetition, order matters, from $\{ -9, -8, \ldots ,0, \ldots ,8,9 \}$.
- How many ways are there to do this when the sum of the last $3$ numbers are even?
- How many ways are there to do this when exactly three numbers chosen are multiples of $3$?
In (Q1.), I got $9^3+3 \cdot 9 \cdot 10^2$ total ways to choose the last three numbers (by counting the number of ways a sum of three ordered numbers can be even), and so my answer is $19^7 \cdot (9^3+3 \cdot 9 \cdot 10^2)$.
In (Q2.), I arrived at the answer: $12^7 \cdot 7^3 \cdot {8 \choose 3}$ (first choose the $7$ numbers that aren't multiples of $3$, then choose the multiples of 3 and also locations to insert them twixt the first $7$).
I hope this is considered a legitimate question! Thank you.