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So I can not seem to get my head around this and could not make sense of online calculators. I think what I'm trying to figure out should be pretty simple.

I have $1$ thing, and $10$ different ways that thing can be. So that's obviously $10$. Now say I add one thing, so now there are $2$ things. Each of which has $10$ different ways that thing could be. So I would just multiply that right? And now with $2$ things, there would be $100$ combinations? Say I add a third thing. That would be $1000$?

That is the way I saw it in my head. Just $10$ increasing exponentially with each added "thing" so say there were $20$ "things" it would be $10^{20}$. Am I right? I am horrible at math, and that just seems almost too simple and not correct.

Vinyl_cape_jawa
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Justan Kellogg
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    This is correct – JuliusL33t Feb 23 '19 at 17:58
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    Yes, if I understand what you mean by "the number of ways that thing can be," you are correct. – saulspatz Feb 23 '19 at 17:58
  • You have $10$ choices for each type of object, so you indeed have $10^3 = 1000$ choices for selecting one object of each of the three types. To convince yourself, try working with smaller numbers. Say you wish to choose one book and one DVD from two different books and three different DVDs. If you draw a tree diagram that begins with your choice of book, you will have three choices of DVD for each book, giving you a total of $2 \cdot 3 = 6$ ways to select one book and one DVD. – N. F. Taussig Feb 23 '19 at 18:05
  • Yes, you're right. And in this case Thing $1$ could be the first digit of a number, Thing $2$ could be its second digit and Thing $3$ could be its third digit—making $1000$ different $3$-digit numbers $000$ to $999$, as it should. – timtfj Feb 23 '19 at 19:31

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Thanks for the input guys! I thought that was correct, I was just over thinking and making myself think I was wrong!

Justan Kellogg
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