# find the number of possible 5-card hands taken from a standard 52-card deck that contain 2 aces and 3 cards th?

3 cards that are not aces.

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• Jared
Lv 7

There are FOUR Aces.

You MUST choose 3 of the 4. You can do this as follows:

(4 c 3) ......."four choose three"

(which you should immediately know equals 4)

Now how many cards are left to choose from?

NO aces, therefore there are:

52 - 4 = 48 cards left.

Therefore you just choose 2 from the rest (this one you will NOT know immediately...of course it's big):

(48 c 2) = 1128

Therefore the number of hands (distinct hands) should be:

4 * 1128 = 4512

This makes the probability of this type of hand to be:

Total number of hands:

(52 c 5) = 2,598,960

-->

p = 4512 / 2,598,960 = .001736 = .1736%

EDIT:

WOW, I did that totally wrong, it should be 2 aces and 3 non-aces...doesn't really change much here's the correct one quickly:

(4 c 2) * (48 c 3) = 6 * 17,296 = 103,776

Of course this should be an "easier" hand to get (rather than 3 aces like I did above) and sure enough there are MORE hands that satisfy this property.

The probability of getting THIS hand would be:

103,776 / 2,598,960 = .03993 = 4% That's almost 1 in 20.