From a standard pack of 52 cards, three are selected at random. Find the probability that they are the jack of spades, the two of clubs and the seven of diamonds.
Using combinatorics: $\frac1{52C3}$ gets you the answer
Using probability, shouldn't $\frac1{52} \times \frac1{51} \times \frac1{50}$ get you the same answer?
Why isn't this the case?
Using combinatorics -
$\frac1{52C3} = \frac{1}{\frac{52!}{49! * 3!}}$
= $\frac{1}{\frac{52 * 51 * 50 * 49!}{49! * 3!}}$
=$\frac{3!}{52 * 51 * 50}$
Using probability -
$\frac{1}{52} * \frac{1}{51} * \frac{1}{50}$
But 3 cards arrange in 3! ways.
So we have,
3! * $\frac{1}{52} * \frac{1}{51} * \frac{1}{50}$
=$\frac{3!}{52 * 51 * 50}$
In combinations you don't need to care about arrangements.
Hint: You can choose the jack of spades ($J$), the two of clubs ($C$) and the seven of diamonds ($7$) in $3!=6$ ways:
$JC7,J7C,CJ7, C7J, 7CJ, 7JC$
Therefore $\frac1{\binom{52}{3}}=6\times \frac1{52} \times \frac1{51} \times \frac1{50}=\frac{1}{22100}$
The reason your two answers are different from each-other is because you are misunderstanding how to represent a marginal probability. If you know about independence in probability, then the card draws are dependent on each other and therefore the solution cannot be drawn as the product of the individual probabilities.
The key to problems of this form is to identity the number of possible outcomes $n$ and the number of desirable outcomes $N$.
Here it is clear with the principle of multiplication that $$n=3!$$ and $$N=\frac{52!}{49!} = 52\times 51\times 50$$
Now, $$P(A) = \frac{n}{N} = \frac{3!}{52\times 51\times 50}$$
Because the three draws are dependent, we cannot simply multiply the probabilities. This can be shown by the definition of marginal probability. That is why you are getting the wrong answer.
If this doesn't make sense, convince yourself that this logic is correct and therefore your logic is incorrect.