How Many Combinations

1st Question:

So the pick 3 lottery thing, you can pick a 3 digit number such as 0-3-1, or 3-9-5, or even 4-5-5, etc.

I was wondering how many total combinations you could make, so I did the whole 10 times 10 times 10 thing I learned in Prob and Stat class and got 1000 total combos.

But people say it is wrong, and that there are more than 300 combos.

So how many combinations can you make?

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2nd Question:

Also, if you do end up picking the winning number (such as picking 3-4-5, and 3-4-5 is indeed drawn) then you win 500 dollars. If the numbers you picked are drawn, but not in sequence (such as picking 3-4-5, but the drawn number is 4-5-3) then you win 40 dollars.

With each ticket being 50 cent, IF there is 1000 combos, and you bought all 1000 combos for a total of 500 dollars, wouldn't you win back your 500 dollars plus 200 dollars for also choosing the other 5 numbers (5 times 40 is 200), then you would win 700 dollars. So you would make a profit of 200 dollars right?

Q1. 1000 is correct. Just think of the three digit numbers that are possible: 0 (001), 2, 3, etc up to 997, 998, 999. That's 1000 numbers.

Q2. Let's assume the numbers that come up are 345 as you suggested. And lets suggest you bought all 1000 numbers (BTW, they will cost you $1 each, that's $1,000). For having 345 you would win $500. You would also own the "anagrams" of 345 (354, 435, 453, 534 and 543). There are just five of these, so you would win an additional 5x40 = $200, making a total of $700 for your $1,000 outlay.

Your error was assuming the cost of a ticket was 50cents, and not $1.

Q1. Yes, 1000 combinations is correct by the fundamental counting principle: there are 10 possible values of the first digit, 10 possible values for the second digit, and 10 possible values of the third digit; 10 * 10 * 10 = 1,000.