1. ## Combinations/Permutations Card problems

You have a standard deck of 52 cards. How many possibilities are there if your turn 6 cards over one by one in the order: an ace, a king, an Ace, a King, an Ace, and a King?

I am trying to help my friend out with his homework but its been so long since I have messed with permutations, combinations, and probability that I am blanking on remembering how to do it. Any help would be greatly appreciate for us. Thanks.

2. ## Re: Combinations/Permutations Card problems

Originally Posted by Tokenfreak
You have a standard deck of 52 cards. How many possibilities are there if your turn 6 cards over one by one in the order: an ace, a king, an Ace, a King, an Ace, and a King?
$\displaystyle (4)(4)(3)(3)(2)(2)$

3. ## Re: Combinations/Permutations Card problems

Originally Posted by Plato
$\displaystyle (4)(4)(3)(3)(2)(2)$
Ok, well that makes sense now since I have use that one king and ace and can't use that one again so it is one less each time. Thanks for that, helped out a lot. We have a ton more of these too do but we just need help on these next two and we should be able to do the rest since they are similar.

You have a standard deck of 52 cards. How many possibilities are there if your turn 6 cards over one by one in the order: an ace, a king, an Ace, a King, an Ace, and a King and each next king is the same suit as the ace before it?

You have a standard deck of 52 cards. How many possibilities are there if your turn 6 cards over one by one in the order: an ace, a king, an Ace, a King, an Ace, and a King and at least one of the kings is NOT the same suit as the ace before it?

4. ## Re: Combinations/Permutations Card problems

Originally Posted by Tokenfreak
You have a standard deck of 52 cards. How many possibilities are there if your turn 6 cards over one by one in the order: an ace, a king, an Ace, a King, an Ace, and a King and each next king is the same suit as the ace before it?
$\displaystyle (4)(1)(3)(1)(2)(1)$

Originally Posted by Tokenfreak
You have a standard deck of 52 cards. How many possibilities are there if your turn 6 cards over one by one in the order: an ace, a king, an Ace, a King, an Ace, and a King and at least one of the kings is NOT the same suit as the ace before it?
$\displaystyle (4)(4)(3)(3)(2)(2)-(4)(1)(3)(1)(2)(1)$

5. ## Re: Combinations/Permutations Card problems

Ok, thanks that help us out a lot. We have now successfully finished all the problems with your help except for one last one that is stumping us.

You have a standard deck of 52 cards. You pick a card, look at it, and return it back to the deck. You pick a card 6 times, but you have a bad memory and the only thing you remember are the suits of the picked cards, not even the order. How many possibilities are there?

6. ## Re: Combinations/Permutations Card problems

Originally Posted by Tokenfreak
You have a standard deck of 52 cards. You pick a card, look at it, and return it back to the deck. You pick a card 6 times, but you have a bad memory and the only thing you remember are the suits of the picked cards, not even the order. How many possibilities are there?
I don't understand what is asked.
How many possibilities are there? For what?

7. ## Re: Combinations/Permutations Card problems

Originally Posted by Plato
I don't understand what is asked.
How many possibilities are there? For what?
How many possibilities are there that you pick a card 6 times, look at it, and return it back to the deck; but you only remember the suits of the picked cards and not even the order.

Thats the best I can rephrase it lol as I don't really understand what I am looking for either.

8. ## Re: Combinations/Permutations Card problems

Originally Posted by Tokenfreak
How many possibilities are there that you pick a card 6 times, look at it, and return it back to the deck; but you only remember the suits of the picked cards and not even the order.
Thats the best I can rephrase it lol as I don't really understand what I am looking for either.
Well it seems to be asking for content with respect to suits. Like all spades; three clubs, two diamonds, and heart; or whatever.
If that is correct then the answer is $\displaystyle \binom{6+4-1}{6!}=\frac{9!}{(6!)(3!)}$.
That is count of multi-selections.

9. ## Re: Combinations/Permutations Card problems

Thanks again.

,

,

### cards permut

Click on a term to search for related topics.