Math question.

[evans075]

Two questions that a teacher of mine asked in class but didn't have time to explain the answers to. I've been racking my brain on how to work these two questions. So could someone explain how you would go about doing this!!!

How many different 8 card hands are there that contain exactly 2 suites with 4 cards from each suite?

How many different 8 card hands are there that contain only face cards?

[theflyingcow]

Quick question that pertains more to the 2nd one, do suits matter, like a hand containing the A[img]/images/graemlins/club.gif[/img] would be different from an identical one containing the A[img]/images/graemlins/diamond.gif[/img]? E.g. AQQQ.

[evans075]

Yes one containing A [img]/images/graemlins/club.gif[/img] would be different from one containing A [img]/images/graemlins/diamond.gif[/img] in AQQQ

[jesse8888]

C(X Y) is going to be used to mean X choose Y. I think these are right, but would appreciate double checking.

Second question: C(12 8) or (12!)/((8!)*(4!)) = 495
There are 12 face cards, you must choose 8.

First question: C(4 2) * C(13 4) = 4290
There are 4 suits, you must choose 2, and there are 13 cards of which you must choose 4.

[jogsxyz]

1. 13C2 * 13C2 * 13C2 * 13C2
In each of the 4 suits there are 13 cards. You choose 2 from each of the suits.
52C8
You choose 8 from the 52 card deck.
13C2 = 13*12/(2*1) = 78
52C8 = 52*51*...45/8! = whatever
78*78*78*78
guess you don't need the denominator.
you calculate it.

[alThor]

[ QUOTE ]
C(X Y) is going to be used to mean X choose Y. I think these are right, but would appreciate double checking.

Second question: C(12 8) or (12!)/((8!)*(4!)) = 495
There are 12 face cards, you must choose 8.

First question: C(4 2) * C(13 4) = 4290
There are 4 suits, you must choose 2, and there are 13 cards of which you must choose 4.

[/ QUOTE ]

Did you mean:
First question: C(4 2) * C(13 4) * C(13 4)

[filsteal]

[ QUOTE ]
[ QUOTE ]
C(X Y) is going to be used to mean X choose Y. I think these are right, but would appreciate double checking.

Second question: C(12 8) or (12!)/((8!)*(4!)) = 495
There are 12 face cards, you must choose 8.

First question: C(4 2) * C(13 4) = 4290
There are 4 suits, you must choose 2, and there are 13 cards of which you must choose 4.

[/ QUOTE ]

Did you mean:
First question: C(4 2) * C(13 4) * C(13 4)

[/ QUOTE ]

I hope he did, because your answer is correct. You need a C(13 4) for EACH suit.

[jogsxyz]

[ QUOTE ]
[

Did you mean:
First question: C(4 2) * C(13 4) * C(13 4)

[/ QUOTE ]

Misread the question. That's the answer to question 1.