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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 $\{ -9, -8, \ldots ,0, \ldots ,8,9 \}$.

  1. How many ways are there to do this when the sum of the last $3$ numbers are even?
  2. 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.

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This is almost perfect, but to place 3 numbers among the 7 others, you have to choose 3 locations among 10.

So your last factor is in fact $\binom {10} 3$, not $\binom 8 3$. Everything else is fine.

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  • $\begingroup$ Ah, of course. Thank you :). $\endgroup$ – donqui Oct 29 '11 at 15:59

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