Probability-Deck of cards

Question

I know these questions have already been discussed. I found these questions in Advanced Maths Forum and want to discuss them once more-

1) A club has a membership of 9 men and 12 women...21 people total. A 5-member committee is to be selected to attend a conference. How many committes with no more than three women be formed? 

2) How many ways we can have 5 cards in a hand from a deck of cards with 3-of-a-kind (and only 3-of-a-kind -- the other two cards cannot match the first 3, nor each other).

I know the answers. So pls provide me with the explaination and logic.

Thanks,
Vijo.

banerjeea_98 · Answer

1. Total ways - gps with 4- gps with 5 women

21C5 - 12C4*9C1 - 12C5



2. Don't understand this...hand of how many cards ? R there 5 cards in the hand ?

Vijo · Answer

Yaa there are 5 cards in hand in the second question.

MA · Answer

Alternatively, (9C2x12C3)+(9C3x12C2)+(9C4x12C1)+(9C5)

swath20 · Answer

1. 21c5- 12c5-9c1*12c4


2. ???????

HongHu · Answer

First you choose from one number of the 13, then you choose three from the four cards of the same kind: C(13,1)*C(4,3)
Then you choose the rest two cards from the rest 12 numbers, then you choose one card from each of the two kinds: C(12,2)*C(4,1)*C(4,1)

Multiply them together.

vprabhala · Answer

i am still unclear on the 2nd one..

ketanm · Answer

First I will try to clear the question :

We need to pick five cards from the deck of 52 cards. Out of these five three cards should be of same kind. This means that their face values should be same. And the rest two cards have to be different. Also the rest two cards can not have same face value. What is the number of ways can this be choosen?

Answer: 

1) The first thing is to choose one number from the 13 (Ace=1 to 
 King=13). Lets say King. C(13,1)
2) Choose three Kings from the four Kings = C(4,3) 
3) Now we need to choose the rest two cards from the rest 12 numbers 
  , 
4) Choose two kinds (since they both can not be of same kind)...e.g 
 choose Queen and Ten (say) C(12,2)
5) Then choose one card from each of the two kinds: C(4,1) for each 


Hence number of ways = C(13,1) * C(4,3) * C(12,2) * C(4,1) * C(4,1)

Ketan

vprabhala · Answer

that makes sense..
thanks ketan

Rupstar · Answer

21C4- (12c4*9c1+12c5)
i got that but shouldnt the second one be

13c1*4c3*(12c1*4c1)*(11c1*4c1)
or is that the same thing?
that was a good question!

Vijo · Answer

Exactly, I landed with the same answer as Rupstar's

13c1*4c3*(12c1*4c1)*(11c1*4c1) 

but it is not same as Hong's answer

Hong's answer is right. But why not this logic? Where was i going wrong???? Hong, plz explain...

HongHu · Answer

Yours is ordered for the last two cards. In other words, in your approach if I picked Q and then 10, you count it as different from if I picked 10 and then Q. If you divide your answer by two you should get the same answer as mine.

Rupstar · Answer

OKAY 
so what we were doing was ordering it then i get it now thanks Hong!

Probability-Deck of cards

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