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Five cards are drawn from a standard deck (not replaced). Determine the probability of drawing exactly 3 hearts and 2 diamonds.

The expression for the probability is:

$\frac{\binom{13}{3}\binom{13}{2}}{\binom{52}{5}}=\frac{143}{16660}$

Then, I used an another way to do it, by multiplying the probability of drawing the card at each draw.

$\overbrace{\frac{13}{52}\frac{12}{51}\frac{11}{50}}^{\mbox{hearts}} \overbrace{\frac{13}{49}\frac{12}{48}}^{\mbox{diamonds}}$

Then the math teacher corrected me, she added the the number of permutations of 5 cards of that type.

$\frac{13}{52}\frac{12}{51}\frac{11}{50} \frac{13}{49}\frac{12}{48}(\frac{5!}{3!2!})$

I don't know why that term for number of permutations is required for that expression, and why that works.

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    You calculated the probability of drawing a heart on the first three draws followed by drawing diamonds on the fourth and fifth draws. What you are _asked_ to find is the probability of drawing three hearts and two diamonds **in any order**, not just three hearts first and then two diamonds. The number of **combinations** of three hearts and two diamonds is $\binom{5}{2}$ (pick the two places where diamonds will occur out of the five places).2012-05-22

2 Answers 2

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In your expression, you have given the probability that the first three cards are hearts and the final two are diamonds (♥♥♥♦♦).

However, that isn't the only way to get a hand of three hearts and five dimaonds. Instead you could have ♦♦♥♥♥ or ♥♦♥♦♥ or a number of other combinations.

How many combinations are there? From five cards, you are choosing three of them to be hearts, so there are ${5\choose 3} = 5!/(3!2!)$ combinations. This is the multiplicative factor that you missed out, and your teacher added in.

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    10k is just around the corner... (drums are rolling in the background)2012-05-23
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The same 5 cards can be arranged in 10 different ways.