From a deck of 52 cards, three cards are drawn at random.…

From a deck of $52$ cards, three cards are drawn at random. Find the probability that these three cards will be a three, a seven and an ace.

Step 1: All outcomes

\[ \binom{52}{3} = \frac{52\cdot51\cdot50}{6} = 22\,100. \]

Step 2: Favourable outcomes

The deck has four threes, four sevens and four aces. Choose one of each:

\[ 4\cdot4\cdot4 = 64. \]

Step 3: The probability

\[ \boxed{P = \frac{64}{22\,100} = \frac{16}{5525} \approx 0.0029.} \]

That is roughly $1$ chance in $345$.

Step 4: Check by drawing in order

Draw the cards one by one. The first card must be a three, seven or ace: $\frac{12}{52}$. The second must be one of the two remaining ranks: $\frac{8}{51}$. The third must be the last rank: $\frac{4}{50}$.

\[ P = \frac{12}{52}\cdot\frac{8}{51}\cdot\frac{4}{50} = \frac{384}{132\,600} = \frac{16}{5525}. \ \checkmark \]

Step 5: A common mistake

$\frac{4}{52}\cdot\frac{4}{51}\cdot\frac{4}{50}$ is the probability of drawing them in the order three, seven, ace. Since any of the $3!=6$ orders is acceptable, multiply by $6$: $6\cdot\frac{64}{132\,600}=\frac{16}{5525}$.