In how many ways can three aces be drawn?

[YODA0311]

probability help pleasezz

1. In how many ways can three aces be drawn?



2. I used this formula - P= n!/(n-r)!
n r


3. Here is my attempt- 52!/(52-3)!=52!/49!= 132600, but that is off to me.

I googled this problem and the person got "There are 4*3*2 = 24 sequences in which 3 aces can be drawn from a deck containing 4 aces." However I do not know how they reached their conclusion :(

 

[YODA0311]

well i think i got it, however i would like to know if it is correct
n!/(n-r)!=4!/(4-3)!=4!/1=24/1=24 possible ways to draw three aces

 

[Dick]

well i think i got it, however i would like to know if it is correct
n!/(n-r)!=4!/(4-3)!=4!/1=24/1=24 possible ways to draw three aces

That's right if you are considering order, i.e. you consider clubs-diamonds-spades to be different from clubs-spades-diamonds.

 

[Redbelly98]

However, I will add that in card-drawing problems, one typically does not consider the order to be important. I.e., clubs-diamonds-spades and clubs-spades-diamonds are considered to be the same.

While there is a formula that gives the right answer, the answer to this one can easily be found (or checked) using common sense.

Moderator's note: thread moved from Intro Physics to Precalc Math.