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How many different ways are there to choose 2 of 4 different colors?

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If you could explain as well, that would be great! :D
For your first pick, you can pick any of 4 colors, so that is going to be "4 times something" so far. That leaves 3 colors of which you can pick any of the three. So, you have "4 times 3" so far. Now, you in effect have "double-counted" because the order doesn't matter. So, you have "4 times 3 divided by 2" for your answer. This is 4C2 which is 4!/(2! times 2!)
The answer is 6?

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btw, "!" notation, the exclamation mark, is "factorial", which is a number times every successive integer under it until you get to 1. Yes, the answer is 6.
The answer is 6 as long as it doesn't matter which of the 2 colors was chosen first. Then the answer would be 12.
Okay I see now! Thank you so much. You have helped a ton! :D
Good luck to you in all of your studies and thx for the recognition! @HorseCrazyGirlForever And you're welcome!
A good way to visualize that it is 6 and that there is no order to the picking is to imagine grabbing 2 at the same time blindly out of a bag.
Sounded funny. Visualizing blindly.

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