Probability of getting exactly 3 aces from drawing 7 cards by two methods This is the problem number 4 from Tutorial question in 6.041/6.431: Probabilistic Systems Analysis from MIT OCW. Basically you are dealt 7 cards from well shuffled 52 cards deck, and asked to find the probability of getting exactly 3 aces.
One way I approached the problem is by multiplying the possible number of aces (4 choose 3) by the probability of getting 4 non-aces and 3 aces: P(A) = ${4\choose 3}$ $\frac{48}{52}$ $\frac{47}{51}$ $\frac{46}{50}$ $\frac{45}{49}$ $\frac{4}{48}$ $\frac{3}{47}$ $\frac{2}{46}$ $\approx$ $\mathrm{6.65}\!\cdot\!\mathrm{\mathrm{10}^{-4}}$
Now, if I'm doing this by counting the possible number of ways of getting exactly 3 aces in 7 cards and dividing it with the possible number of ways of having 7 cards however, I get different result :
P(A) = $\frac{{4\choose 3}{48\choose 4}}{52\choose7}$ $\approx$ $\mathrm{5.82}\!\cdot\!\mathrm{\mathrm{10}^{-3}}$
Can someone tell me in which part I am wrong?
 A: 
One way I approached the problem is by multiplying the possible number of aces (4 choose 3) by the probability of getting 4 non-aces and 3 aces: 

Don't multiply by the possible number of aces.   You are already accounting for that.
Rather, multiply by the possible number of positions for the aces among the seven cards.
$$\binom 7 3\dfrac{48\cdot 47\cdot 46\cdot 45\cdot ~~4\cdot ~~3\cdot ~~2}{52\cdot51\cdot 50\cdot 49\cdot 48\cdot 47\cdot 46}=\dfrac{7!}{4!~3!}\dfrac{\dfrac{48!}{44!}\dfrac{4!}{1!}}{\dfrac{52!}{45!}}=\dfrac{\dfrac{48!}{4!~44!}\dfrac{4!}{3!~1!}}{\dfrac{52!}{7!~45!}}=\dfrac{\dbinom {48}{4}\dbinom {4}{3}}{\dbinom {52}{7}}$$

$\tfrac{48\cdot 47\cdot 46\cdot 45\cdot ~~4\cdot ~~3\cdot ~~2}{52\cdot51\cdot 50\cdot 49\cdot 48\cdot 47\cdot 46}$ is the probability for obtaining four consecutive non-aces and then three consecutive aces when dealing cards out in order.
$\tbinom 73$ counts the permutations for 4 non-ace places and 3 ace places.

An alternate method is to evaluate the probability for three places for the aces to be among the first seven places in the deck (and one place among the later forty-five), when the four aces may be placed anywhere among the fifty two places without bias.$$\dfrac{\dbinom 73\dbinom {45}1}{\dbinom{52}{4}}$$
