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I am struggling on a problem like this on my homework.

Suppose 2 quarters, 3 dimes, 6 nickels, and 10 pennies.

If you were to draw four coins, without replacing them, what is the probability of drawing one of each coin?

I am mostly struggling with the fact that isn't the probability different depending on the order of how you pull the coins? I think I am missing some of the concept here mostly.

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HINT: You’re not concerned with order in this problem: without replacement just says that you’re drawing a set of four distinct coins. You might as well think of drawing them all at once. There are $21$ coins, so there are $\binom{21}4$ different sets of four coins that you could draw, all equally likely (unless, of course, you deliberately pay attention to the different sizes of the coins!). How many of those sets contain exactly one coin of each denomination?

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    @Tom: That’s right. By the way, the multiplication rule for counting the ways to make a series of choices (also sometimes called the Chinese menu rule) comes up all the time, so you should try to get comfortable with it as soon as possible.2012-12-12
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Give the coins individualities, by putting identity numbers on them.

There are $\dbinom{21}{4}$ ways of choosing $4$ coins. All these ways are equally likely.

There are $(2)(3)(6)(10)$ ways to choose a coin of each kind.

Divide.