How many ways can 5 cards be selected from a 52 card deck?

Get the answer to your homework problem.

Try Numerade free for 7 days

Michigan State University

Sara U.

Calculus 3

1 year, 2 months ago

We don’t have your requested question, but here is a suggested video that might help.

A standard deck of playing cards has 52 cards. a. How many different 5 -card poker hands could be formed from a standard deck? b. How many different 13 -card bridge hands could be formed? c. How can you tell that numbers of combinations are being asked for, not numbers of permutations?

Check my profile and contact me for one on one online tutoring.

If you take all the possible 5 card draws, and multiply that by the probability all 5 cards are hearts, the result is all the possible 5 heart draws.

The number of ways to draw 5 cards from 52 is a combination...52C5. Remember the formula for combinations:

nCk = n! / (k! * (n-k)!. In this case n=52 and k=5:

52!/(5!*(52-5)!)= 52!/(5!*47!) = (52*51*50*49*48)/(5*4*3*2*1) = 2,598,960.

That's the total number of five card combinations. But what is the probability that all five cards are hearts? There are 13 hearts in the deck of 52 cards.

P card 1 being a heart is 13/52. P card 2 being a heart (there's one less card, and one less heart) is 12/51 P card 3: 11/50 P card 4: 10/49

P card 5: 9/48

The probability of all five events is the product of their probabilities:

13/52 * 12/51 * 11/50 * 10/49 * 9/48 = 0.000495198079231693.

The number of combinations of five cards, times the probability of all five hearts, gives us the answer we want.

0.000495198079231693*2,598,960=1,287.

The question is:

In a game of poker, five players are each dealt $5$ cards from a $52$-card deck. How many ways are there to deal the cards?

The answer is $\frac{52P5}{(5!)^5}$ or $\frac{52!}{47!5!}$ I understand the reason after seeing the answer: $25$ ordered cards are picked from the deck of $52$, then divided by the number of ways each hand can be ordered, because the order of a hand does not matter.

However, before seeing the answer I came up with this attempt:

You choose $5$ cards from $52$, the order doesn't matter so we have $52\choose{5}$. Then you're left with $47$ cards and for the next hand we have $47\choose{5}$. The final sum is ${{52}\choose{5}} + {{47}\choose{5}} + {{42}\choose{5}} + {{37}\choose{5}} + {{32}\choose{5}}$.

This sum is considerably larger, but while I understand the reason $\frac{52!}{47!5!}$ is the correct, I can't find the mistake in my first attempt.