# Probability of drawing a combination of coins?

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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Ok, I feel bad here, but I also for some reason can't think of how to figure out how many of those sets have one of each coin. I'm usually pretty decent at math but recently this class has been kicking my butt, although I think it is because it is online and there is no good materials. – Tom Dec 12 '12 at 0:43
@Tom: It’s just a problem in counting the ways to make a series of choices. There are $2$ ways to choose the quarter, $3$ ways to choose the dime, $6$ ways to choose the nickel, and $10$ ways to choose the penny. These choices are made independently, so they can be combined in all possible ways: $2\cdot3\cdot6\cdot10$. – Brian M. Scott Dec 12 '12 at 0:45
Ok, wait, so am I getting this right now, there is 21 ways to draw each coin, and because there is 21 coins and you are drawing 4 there will be 21C4 total combinations, so it would be 21/5985. Or am I way off base? – Tom Dec 12 '12 at 0:50
@Tom: Off base, I’m afraid. There are just $2$ ways to draw a quarter. There are $3$ ways to draw a dime, so there are $2\cdot3=6$ different (equally likely) combinations of one quarter and one dime that you could draw. For each of those $6$ there are $6$ different nickels that you could draw, so there are $2\cdot3\cdot6=36$ different (equally likely) combinations of one quarter, one dime, and one nickel. Can you finish it now? – Brian M. Scott Dec 12 '12 at 0:53
Ok, so it would be 2 \cdot 3 \cdot 6 \cdot 10=360, and then over 21C4? – Tom Dec 12 '12 at 0:56

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.

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