unordered selection with repetition (1 Viewer)

[mojako]

Is this covered in the HSC course?
I remembered thinking about it and couldnt find a working formula / method.

example:
2 cards are chosen at random from a collection of 5 cards, labelled A, B, C, A, D.
In how many way can they be chosen?

That can be done fairly quickly without any maths. How about when there are 198 cards and we're chosing 31 cards, and 17 of the cards are labelled A while the rest are all distinct to one another?

EDIT: sorry, this is actually unordered selection with identical elements.
I have another Q re repetition on post #4
Thanks.

 

[turtle_2468]

Is this covered in the HSC course?
I remembered thinking about it and couldnt find a working formula / method.

example:
2 cards are chosen at random from a collection of 5 cards, labelled A, B, C, A, D.
In how many way can they be chosen?

That can be done fairly quickly without any maths. How about when there are 198 cards and we're chosing 31 cards, and 17 of the cards are labelled A while the rest are all distinct to one another?

Just thinking on the run, but...
you could always do a summation thing.. ie case 1: none of the 31 cards are A's. case 2: there's 1 A. Then for each case, you've determined how many non-A cards you need to choose (eg 30 non-A's for case 2) then just 198C30...
Actually, so then generalising, the answer is
198C31+198C30+...+198C14. (14 is for the case when 17 cards - the maximum - are A's)

 

[mojako]

what about.. when 17 of the cards are A, 26 of the cards are B, 13 of the cards are C, and the rest are distinct to one another?
the summation gets more complicated

 

[mojako]

sorry.. the title of the thread should be "unordered selection with identical elements"

But now I also have a question re unordered selection with repetition.
Example:
5 cards are drawn from a standard pack of 52 cards. Each card is replaced as soon as it is picked up.
In how many combinations can we have the 5 cards?

Thanks.

 

[Euler]

not sure about a general formula for the original problem...

for the 52 cards one, is it 52^5?

 

[mojako]

not sure about a general formula for the original problem...

for the 52 cards one, is it 52^5?

I think that's if it is ordered.

Here's the types of situations I've gathered:
1. order is important
1.1. things (eg cards) are replaced as soon as each one is taken out: use n^k
1.2. things are not replaced: use nPk
2. order is important
2.1. things are replaced: ??????
2.2. things are not replaced: use nCk

Also, when there are identical elements (eg 2 cards labelled A and 3 cards labelled B) with
1.1 : ???
1.2 : ??? (except in the specific case where we take all the cards: divide the nPn by (a! * b!), where a=2 and b=3 in the example above)
2.1 : ???
2.2 : use the summation explained by turtle

If that makes sence