# Why is this term required for a probability about drawing cards from a deck

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). –  Dilip Sarwate May 22 '12 at 19:09

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) –  mixedmath May 23 '12 at 8:16