Permutations and combinations (1 Viewer)

[Petinga]

A standard pack of 52 cards consists of 13 cards of each of the four suits: spades, hearts, clubs and diamonds.

a) In how many ways can six cards be selected without replacemnt so that exactly two are spades and four are clubs? (assume that order of selection of six cards is not importnat)

b) In how many ways can six cards be selected without replacemnt if at least five must be of the same suit?(assume that order of selection of six cards is not importnat)

help appreciated asap

 

[Templar]

13C2*13C4

4*13C5*47

 

[Mountain.Dew]

ill offer a bit of explanation...

(a). 13 spades and 13 clubs. can only choose 2 spades, so 13C2. need to have 4 clubs, so 13C4. Then, multiply to get 13C2*13C4

(b) just consider one suit. must have at least 5. so, there is option of 5 cards same, plus one OTHER. other option is 6 cards same suit.

SO, for at least 5 cards: 13C5 * 39 * 4 (there are 39 cards that arent the same suit of the 5 selected. we also times by 4 because there is 4 suits.

AND for the 6 cards, its simply 13C6 * 4 (4 suits)

final answer is: 13C5 * 39 * 4 + 13C6 * 4 for (b)

 

[Templar]

There are 39 cards that aren't in the suit selected.

 

[Mountain.Dew]

There are 39 cards that aren't in the suit selected.

oops my bad...i included the jokers in the deck XD

is now changed to 39

 

[airie]

combinatorics. me least fave topic :S

i have a question too: in how many ways can you sort seven people into four non-empty teams? the teams are indistinguishable but the people are. i mean i just always get confused over this question and it just all becomes a mess in the end :S

 

[Templar]

Possible team formations:

1:1:1:4 7C4
1:1:2:3 7C3*4C2
1:2:2:2 7C2*5C2*3C2

Add them all up.

 

[airie]

Possible team formations:

1:1:1:4 7C4
1:1:2:3 7C3*4C2
1:2:2:2 7C2*5C2*3C2

Add them all up.

so that gives...wait...i can't put two and two's together...875. o.0 are you sure? i have the answer here and it says 350 ways. but problem is, i don't know how they got that answer :S could you check if there were any errors in your solution please? thanks

 

[Riviet]

so that gives...wait...i can't put two and two's together...875. o.0 are you sure? i have the answer here and it says 350 ways. but problem is, i don't know how they got that answer :S could you check if there were any errors in your solution please? thanks

I added them up too and also came up with 875, looked at Templar's solution, and by inspection it looks fine at the moment.

If I figure out why it's wrong, I'll post it up.

EDIT! I just figured out what the problem is. For the 2 : 2 : 2 : 1 combination, let the respective teams be A : B : C : D. We start with 7 people to pick for A. We move on to B with 5 to go since 2 have already been assigned to A. Then we move on to C with 3 left, and finally to D with the only last one. So we multiply 7x5x3x1 because these happen at the same time.

So number of ways= ⁷C₄ + ⁷C₃.⁴C₂ + (7x5x3x1) = 350.

Riv.

 

[icycloud]

Possible team formations:

1:1:1:4 7C4
1:1:2:3 7C3*4C2
1:2:2:2 7C2*5C2*3C2

Add them all up.

You have to divide the last term by 3!. I.e. (7C2 * 5C2 * 3C2) / 3!
Because the teams are of the same size, you inadvertently count the same thing 3! times over so you must divide by 3!.

 

[Riviet]

You have to divide the last term by 3!. I.e. (7C2 * 5C2 * 3C2) / 3!
Because the teams are of the same size, you inadvertently count the same thing 3! times over so you must divide by 3!.

Yeah, that makes sense now, I knew it was the third combo that had problems because of the 2 2 2 being repeated 6 times.

 

[airie]

I added them up too and also came up with 875, looked at Templar's solution, and by inspection it looks fine at the moment.

If I figure out why it's wrong, I'll post it up.

EDIT! I just figured out what the problem is. For the 2 : 2 : 2 : 1 combination, let the respective teams be A : B : C : D. We start with 7 people to pick for A. We move on to B with 5 to go since 2 have already been assigned to A. Then we move on to C with 3 left, and finally to D with the only last one. So we multiply 7x5x3x1 because these happen at the same time.

So number of ways= ⁷C₄ + ⁷C₃.⁴C₂ + (7x5x3x1) = 350.

Riv.

thanx guys! i get it now

...now why does a question always seem so easy after someone has done it? XP