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

Question

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.

Answer 1

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.

Answer 2

The same 5 cards can be arranged in 10 different ways.