Another math question. In how many ways can 5 cards be drawn...

[T_Samner]

without replacement from a deck of 52 so that 3 are kings and 2 are queens and the order of selection matters?

[John Rogers]

192

[lendaddy]

15,625

Something tells me I've forgotten my stats formulas

[mattdavis11]

4/52*3/51*2/50*4/49*3/48



Nevermind, I was showing probablility.

[T_Samner]

No correct answers yet but lendaddy's closest.

[Zero10]

I get 288...
If 15,xxx is closest then I am definitely wrong.

[T_Samner]

Originally posted by Zero10
I get 288...
If 15,xxx is closest then I am definitely wrong.

Oops, sorry I made a mistake. The answer is only a four digit number.

[mikester]

So, you have to figure the permutation of each card.

You have hte permutation of 3 of 4 kings possible in a hand of 5 cards;

P(4,3)

and then the permutation of 2 of 4 queens possible in a hand of 5 cards;

P(4,2)

And the Rule of product says those answers should be multiplied;

P(4,3) * P(4,2)

P(n,m) = n!/(n-m)! or in Excell simply =PERMUT(n,m)

So, we get 24 * 12 = 288.

If order did not matter then it would be a matter of combination and not permutation.

C(4,3) * C(4,2)

C(n,m) = P(n,m)/m! or in Excel =combin(n,m)

4 * 6 = 24 different unordered combinations possible.

Next time do your own math homework.

[T_Samner]

Originally posted by mikester
So, you have to figure the permutation of each card.

You have hte permutation of 3 of 4 kings possible in a hand of 5 cards;

P(4,3)

and then the permutation of 2 of 4 queens possible in a hand of 5 cards;

P(4,2)

And the Rule of product says those answers should be multiplied;

P(4,3) * P(4,2)

P(n,m) = n!/(n-m)! or in Excell simply =PERMUT(n,m)

So, we get 24 * 12 = 288.

If order did not matter then it would be a matter of combination and not permutation.

C(4,3) * C(4,2)

C(n,m) = P(n,m)/m! or in Excel =combin(n,m)

4 * 6 = 24 different unordered combinations possible.

Next time do your own math homework.

Nope. Incorrect.

[Hugh R]

I believe its 5/52! (factoral)+4/51!+3/50!+3/49!+2/48!+1/47! What did I win?

[T_Samner]

A raspberry 'cause that's incorrect.

[Zero10]

Hugh, we are not asking for probability, but rather a # of combinations.....

Well, I feel better since somebody else got 288
I see 4 different kings that can come up in the first place, then 3 kings remain, then 2 kings. I thought by doing it this way it would account for the order of the cards?.....
Ah.... BUT, what if the queens were drawn first?... Is that the catch?
Or what if it went queen-king-queen-king-king?.... I get it now.
there are 288 ways for each configuration, and there are...... 10? ways to arrange it, so 2880 configurations?

Let 1 = queen, 0 = king, here are the ways I come up with.
11000
10100
10010
10001
01100
01010
01001
00110
00101
00011
Did I miss any?

[T_Samner]

Originally posted by Zero10



there are 288 ways for each configuration, and there are...... 10? ways to arrange it, so 2880 configurations?

Winner!



If the numbers 1 through five represent the order in which a card is drawn, then the 3 kings could have been drawn as the first, second and third cards drawn or the first second and fourth cards drawn etc. for a total of C(5,3) = 10 unique orders of drawing a king. For each of these 10 unique groups representing the order in which a king could be drawn there are P(4,3) = 24 ways of assigning the four different kings to a position in the group for a total of C(5,3) x P(4,3) = 240 different possible ways to draw a king. Finally, there are P(4,2) = 12 ways of getting a queen for the 2 remaining cards for a total of C(5,3) x P(4,3) x P(4,2) = 2880.

[Rob Channell]

Probability and Statistics......

Did you here the one about the statistician?
He drowned in a lake with an average depth of only 6 inches.


Geek that I am......I enjoyed the thread.

[mikester]

Originally posted by T_Samner
Winner!



If the numbers 1 through five represent the order in which a card is drawn, then the 3 kings could have been drawn as the first, second and third cards drawn or the first second and fourth cards drawn etc. for a total of C(5,3) = 10 unique orders of drawing a king. For each of these 10 unique groups representing the order in which a king could be drawn there are P(4,3) = 24 ways of assigning the four different kings to a position in the group for a total of C(5,3) x P(4,3) = 240 different possible ways to draw a king. Finally, there are P(4,2) = 12 ways of getting a queen for the 2 remaining cards for a total of C(5,3) x P(4,3) x P(4,2) = 2880.

That doesn't seem correct to me at all; I stand by my answer. I'll send it to my Math prof though and see what he has to say.

[Zero10]

Yay! I win
It was late at night and I was trying to fix a Java program I had written, so I needed a break

[T_Samner]

I like questions like this because it gives the brain something to gnaw on.