I have arrived at a seperate way of arriving at the answer for this probability question.

I am terrible at probability so I am not quite sure if it is a fluke or my logic.

Here's the question:

The Delta Course
Daily GMAT Practice Question
May 5, 2007

The Full House Casino is running a new promotion. Each
person visiting the casino has the opportunity to play the
Trip Aces game. In Trip Aces, a player is randomly dealt
three cards, without replacement, from a deck of 8 cards. If
a player receives 3 aces, they will receive a free trip to
one of 10 vacation destinations. If the deck of 8 cards
contains 3 aces, what is the probability that a player will
win a trip?

A. 1/336
B. 1/120
C. 1/56
D. 1/720
E. 1/1440

The solution they gave is as follows:

The probability of an event A occurring is the number of
outcomes that result in A divided by the total number of
possible outcomes.

There is only one result that results in a win: receiving
three aces.

Since the order of arrangement does not matter, the number
of possible ways to receive 3 cards is a combination
problem.

The number of combinations of n objects taken r at a time is

C(n,r) = n!/(r!(n-r)!)

C(8,3) = 8!/(3!(8-3)!)
C(8,3) = 8!/(3!(5!))
C(8,3) = 40320/(6(120))
C(8,3) = 40320/720
C(8,3) = 56

The number of possible outcomes is 56.

Thus, the probability of being dealt 3 aces is 1/56.

The way I arrived at the solution is :

1) Probability of being dealt an ace out of 8 cards: 3/8
2) Probability of being dealth an ace out of 7 cards: 2/7
3) Probability of being dealt an ace out of 6 cards : 1/6
As cards are not being replaced.

Hence probability of being dealt 3 aces is: 3/8*2/7*1/6 = 1/56.

Your comments are appreciated.