1. ## Combinations

I have a question. The question is:

You are buying 5 postcards from a shop that has 4 different postcards.

1) In how many combinations can you buy 5 cards?
2) How many possible combinations do 2 of the 4 different cards contain?

1) $\left(^{5}\mathcal{C}_4\right)=\frac{5!}{(4!)(5-4)!}= 5$

2) $\left(^{4}\mathcal{C}_2\right)=\frac{4!}{(2!)(4-2)!}=$ 6

Did I do it right or?

Note: These are without repetitions only.

2. I am assuming you made a slight typo in stating the question.

Originally Posted by Keep
You are buying 4 postcards from a shop that has 5 different postcards.

1) In how many combinations can you buy 4 cards?

However, if you really are buying 5 postcards from a shop selling 4 postcards, then it is impossible for the 5 cards to be all different. In this case, assuming that there is no restriction on the number of cards that are the same, the answer is $4^5.$

3. Originally Posted by TheAbstractionist
I am assuming you made a slight typo in stating the question.

However, if you really are buying 5 postcards from a shop selling 4 postcards, then it is impossible for the 5 cards to be all different. In this case, assuming that there is no restriction on the number of cards that are the same, the answer is $4^5.$
Yeah that is how the question is. You are buying 5 cards from a shop that has 4 different postcards. So I get the answer to 1). The answer to 2) is also $4^2$ or?